\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\) Given a set \(S\) and representations of \(S\) on various vector spaces, we can make more representations on other vector spaces in a few ways.
Suppose we have two representations \((V,\rho:S\rar End(V))\) and \((W,\sigma:S\rar End(W))\). We say that a linear map \(\phi:V\rar W\) is an intertwiner from \((V,\rho)\) to \((W,\sigma)\) if for all \(s\in S\), we have that
$$\sigma(s) \circ \phi = \phi \circ \rho(s)$$ In other words, first moving from \(V\) to \(W\) and then acting by \(s\) is the same as first acting by \(s\) on \(V\) and then moving to \(W\).
Intertwiners serve as the appropriate notion of map between representations, since they carry the actions from one representation to the other. You might also see the term \(S\)-equivariant map for the corresponding notion for \(S\)-sets.
Now suppose that \(\phi\) is not just a linear map, but an isomorphism. Then \(V\) and \(W\) are isomorphic as vector spaces, and we can rewrite the intertwiner equation as
$$\sigma(s) = \phi \circ \rho(s) \circ \phi^{-1}$$ which looks a lot like the change-of-basis formula. Since changing basis is really just viewing things from a different perspective, we don't consider it to change anything important. Similarly, we say that if two representations have an intertwiner that's an isomorphism, the two representations are equivalent, differing only in labeling details.
In general we don't distinguish between equivalent representations. So for instance, if we ask for all of the representations of some set, we don't actually mean all representations, but rather for all equivalence classes of representations, or often one representation from each equivalence class.
Now that we've talked about what it means for two representations to be equivalent, let's see how to get inequivalent representations.
Suppose we have a representation \((V,\rho)\), and suppose that for some subspace \(W\subset V\), for all \(w\in W\) and for all \(s\in S\), \(\rho(s)w \in W\). We thus say that \(W\) is a submodule of \(V\), or that \((W,\rho|_W)\) is a subrepresentation of \((V,\rho)\).
Given a module and a submodule, we can form the quotient module, \(V/W\), whose elements are, as described in the post on quotients, equivalence classes of elements of \(V\). We note that we can form an action of \(S\) on \(V/W\), because if \(v\) is equivalent to \(v'\), then \(v-v' \in W\), and so \(\rho(s)v - \rho(s)v' = \rho(s)(v-v') \in W\), and thus \(\rho(s)v\) is equivalent to \(\rho(s)v'\).
We say that \(V\) is a simple module if it has no proper submodules (i.e. submodules that are not all of \(V\) and are not the \(0\) vector space\), i.e. no proper quotient modules; alternatively, we say that \((V,\rho)\) is an irreducible representation.
Now suppose that we have two representations \((V,\rho)\) and \((W,\sigma)\). We can put these together in a few different ways.
Look at the direct sum of \(V\) and \(W\), i.e. \(V \oplus W\) whose elements can be written as \((v, w)\) for \(v \in V\) and \(w \in W\), with the usual addition of \((v,w)+(v',w') = (v+v',w+w')\) and scalar multiplication \(c(v,w) = (cv,cw)\). The corresponding representation is given by
\((V\oplus W, \rho \oplus \sigma)\) where the map \(\rho \oplus \sigma\) acts as
$$(\rho \oplus \sigma)(s)((v,w)) = (\rho(s)v,\sigma(s)w).$$ Translating all of this into matrix form gives that \((\rho\oplus\sigma)(s)\) is a block-diagonal matrix with one block being \(\rho(s)\) and the other being \(\sigma(s)\).
We say that a representation \((U,\pi)\) is decomposable if it is equivalent to \((V\oplus W,\rho\oplus \sigma)\) for some pair of representations \((V,\rho)\) and \((W,\sigma)\) where both \(V\) and \(W\) are not the 0 vector space. A representation that isn't equivalent to a direct sum is thus indecomposable. This is distinct from being simple, as \(V/W\) is not itself a submodule of \((V,\rho)\), although it is sometimes equivalent to one, so modules can fail to be simple but still be indecomposable.
We say that a representation is fully decomposable if it is the direct sum of simple representations. There are some criteria for when we should expect representations to be fully decomposable. One case is if \(S\) is a finite group; another case is if \(S\) is a finite-dimensional simple Lie algebra or Lie group and \(V\) is finite-dimensional. These are the cases I'm mostly concerned with.
Going a bit further, we can also ask how to determine what simple representations a given representation decomposes or reduces into.
Example: Consider the group \(Uni_n(\mb R)\) of upper-triangular \(n\times n\) real-valued matrices with 1s down the diagonal, acting on \(\mb R^n\) in the usual fashion. The vector \((1,0,0,\ldots)\) is sent to itself by everything in this group, and so it is the basis of an invariant 1-dimensional subspace \(V\). However there is no \((n-1)\)-dimensional subspace complementary to \(V\) that is also invariant, since for any \(u\) outside of \(V\), there is a group element \(g\) that sends \(u\) into \(V\). So while \(\mb R^n\) is not simple as a \(Uni_n(\mb R)\) module as it has an invariant subspace, it's indecomposable since we can't find another complementary invariant subspace.
We've also talked about tensor products before, when we talked about coalgebra actions. Recall that for a coalgebra \(C\) and representations \((V,\rho)\) and \((W,\sigma)\), \(C\) acts on \(V\otimes W\) via
$$(\rho\hat\otimes \sigma)(c) = (\rho\otimes \sigma)\Delta(c).$$ We can of course extend this to Hopf algebras. Notably, sets that don't have some sort of comultiplication are not considered to be able to act on tensor products.
Example:
Consider the group \(SO(3)\) acting on \(\mb R^3\) via the usual rotation action. We can examine its action on \(\mb R^3 \otimes \mb R^3\), which as noted above fully decomposes into simple modules. Firstly we note that \(SO(3)\) preserves the symmetric and antisymmetric subspaces of \(\mb R^3 \otimes \mb R^3\), by which we mean the subspaces spanned by elements of the form \(a \otimes b + b \otimes a\) and \(a\otimes b - b \otimes a\) respectively. The antisymmetric subspace is 3-dimensional, and indeed the resulting representation is equivalent to the original representation on \(\mb R^3\).
In the case of the symmetric subspace, if we write the standard basis elements of \(\mb R^3\) as \(a,b,\) and \(c\), we get that the element \(a \otimes a + b \otimes b + c\otimes c\) gets sent to itself by any element of \(SO(3)\), so the vector space spanned by that element is invariant, giving us the trivial representation. So the 6-dimensional symmetric subspace decomposes into a copy of the trivial representation and a 5-dimensional representation, which turns out to also be simple.
Hence as an \(SO(3)\)-module, \(\mb R^3 \otimes \mb R^3\) decomposes into a 1-dimensional, a 3-dimensional, and a 5-dimensional simple module.
Example:
Suppose that we have a cocommutative Hopf algebra \(H\), for instance a group algebra or the universal enveloping algebra of a Lie algebra. Suppose that \(H\) has a module \(V\). Then we can build actions of \(H\) on tensor powers of \(V\), and thus on the entire tensor algebra \(T(V) = \bigotimes^* V\).
Now consider the ideal \(I_S = \left\langle xy - yx\right\rangle\) in the tensor algebra, remembering that we don't use \(\otimes\) to indicate multiplication inside the tensor algebra itself. \(I_S\) is invariant under the action of \(H\) since \(H\) is cocommutative, so we can form the quotient module, \(Sym(V) = T(V)/I_S\), which is the symmetric algebra on \(V\), i.e. the space generated by products of basis vectors of \(V\) where the order of multiplication doesn't matter.
Similarly, given the ideal \(I_A = \left\langle xy + yx \right\rangle\), we can form the quotient module \(Alt(V) = T(V)/I_A\), the alternating or exterior algebra on \(V\) where swapping two vectors in a product gives you a minus sign.
Another example is when \(V\) has a Lie bracket, and thus we can form the ideal \(I_L = \left\langle xy - yx - [x,y]\right\rangle\), which gives us the quotient module \(U(V) = T(V)/I_L\), i.e. the universal enveloping algebra.
Because \(H\) has a coalgebra structure, in all of these cases it preserves the algebraic structures of the quotient modules. How these various algebras decompose into simple or at least simpler \(H\)-modules is of quite some interest to representation theorists. Depending on where \(V\) comes from and the relationship of \(H\) and \(V\), these examples also give interesting results in combinatorics, geometry and theoretical physics.
Monday, February 29, 2016
Monday, February 8, 2016
Lie algebra actions, universal enveloping algebra
\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)Now that we've talked about algebra and coalgebra actions in the context of groups, we can talk about them in the context of Lie algebras.
So let's consider a Lie group \(G\) and its Lie algebra \(\mf g\). Since we'll have several identities floating around, we'll write the identity element of \(G\) as \(e\), whereas identity matrices will be denoted \(I\).
For a vector space \(V\), \(End(V)\) has a Lie bracket, where for matrices \(S\) and \(T\) we have \([S, T] = ST - TS\). A Lie algebra action of \(\mf g\) on a vector space \(V\) is a map \(\rho: \mf g \rar End(V)\) such that the bracket on \(\mf g\) becomes the bracket on \(End(V)\). In other words,
$$\rho([X,Y]) = \rho(X)\rho(Y) - \rho(Y)\rho(X).$$Given an action of \(G\) on a vector space \(V\) with map \(\rho\), we can make a Lie algebra action by saying that for \(X \in \mf g\),
$$\rho(X) = \frac{\rho(e + sX) - \rho(e)}{s}$$This looks like a derivative, and those less comfortable with infinitesimals may replace it with such.
Note that this is not an algebra action, at least not if we use the Lie bracket as the multiplication, because the bracket in the Lie algebra does not become composition in \(End(V)\).
The comultiplication on \(G\) sends \(g\) to \(g \otimes g\). In particular, the identity element \(e\) gets sent to \(e \otimes e\). We view the Lie algebra as elements \(X\) such that \(e + sX\) is in \(G\) for infinitesimal \(s\). So
$$\Delta(e + sX) = (e + sX)\otimes (e + sX) = e \otimes e + sX \otimes e + e \otimes sX$$ where we use the fact that \(s^2 = 0\).
The point of infinitesimals is that they make everything look linear, so we have the idea that
$$\Delta(e + sX) = \Delta(e) + \Delta(sX) = \Delta(e) + s\Delta(X)$$Since as mentioned, \(\Delta(e) = e \otimes e\), we get that \(\Delta(X) = X\otimes e + e \otimes X\). This comultiplication is cocommutative and coassociative.
Thus we get that for two representations \((V,\rho)\) and \((W, \sigma)\), \(\mf g\) acts on \(V\otimes W\) via
$$(\rho \hat \otimes \sigma)(X)(v \otimes w) = \rho(X)v \otimes w + v \otimes \sigma(X)w$$This should look somewhat product-rule-y, in keeping with the understanding of Lie algebras as derivatives of Lie groups.
Although the comultiplication sends \(G\) to \(G\otimes G\), we don't get that the comultiplication sends \(\mf g\) to \(\mf g \otimes \mf g\). Instead it gets sent to \((\mb ke \oplus \mf g) \otimes (\mb ke \oplus \mf g)\). Most of the time we'll leave off the \(e\).
We can make \(\mb k \oplus \mf g\) into a counital coalgebra by defining \(\epsilon(X) = 0\) for all \(X \in \mf g\) and \(\epsilon(c) = c\) for \(c \in \mb k\).
What if we want an algebra that contains the Lie algebra \(\mf g\) where Lie algebra actions of \(\mf g\) become algebra actions? We call such an object the universal enveloping algebra of \(\mf g\), denoted \(U(\mf g)\), and we create it as a quotient.
We start with the tensor algebra \(\bigotimes^* \mf g\), where we're viewing \(\mf g\) as a vector space. Now we impose the Lie algebra structure by looking at the ideal \(I = \left\langle X \otimes Y - Y \otimes X - [X, Y]\right\rangle\) where \([, ]\) is the Lie bracket, and forming the quotient \(U(\mf g) = \bigotimes^* \mf g/I\).
Recall that \(\bigotimes^* V\) has a comultiplication, given by sending \(v \in V\) to \(v \otimes 1 + 1 \otimes v\), and extending by multiplication. It also has a counit, sending \(v\) to \(0\), and an antipode that sends \(v\) to \(-v\). So \(\bigotimes^* V\) is a Hopf algebra. If \(V = \mf g\), then the ideal \(I\) defined above is both an ideal and a coideal and is closed under antipode, so the quotient \(U(\mf g)\) is not only an algebra, but a Hopf algebra.
Given a universal enveloping algebra, we can recover the Lie algebra that it is built from by looking for the elements \(X\) such that \(\Delta(X) = X \otimes 1 + 1 \otimes X\), which we call "primitive" elements. Note that \(\Delta\) respects the Lie bracket, in that
$$\Delta([X, Y]) = [X, Y] \otimes 1 + 1 \otimes [X, Y] = \Delta(X)\Delta(Y) - \Delta(Y)\Delta(X)$$Thus the primitive elements in a Hopf algebra do form a Lie algebra.
For our universal enveloping algebra, we can take a representation of \(\mf g\) and extend it to a representation of \(U(\mf g)\) as follows: any non-scalar element in \(U(\mf g)\) can be written as a sum of products of elements of \(\mf g\), so we focus on such products, writing them as \(X_1X_2\cdots X_k\). Given a representation \((V,\rho)\) of \(\mf g\), we apply \(\rho\) to \(X_1X_2\cdots X_k\) in the obvious fashion, as \(\rho(X_1)\circ \rho(X_2) \circ \cdots \circ \rho(X_k)\). Scalars map also in the obvious fashion, with \(c \in \mb k \subset U(\mf g)\) going to \(cI\in End(V)\). Since \(\rho\) respects the Lie bracket, we get a well-defined algebra representation of \(U(\mf g)\). So now we can use algebra action theorems and techniques to talk about Lie algebra actions.
If we instead look at the ideal \(I_h = \left\langle X \otimes Y - Y\otimes X - h[X,Y]\right\rangle\) for \(h \in \mb k\), we get a slightly different algebra sometimes denoted by \(U_h(\mf g)\). For \(h\) nonzero this is isomorphic to \(U(\mf g)\) just by rescaling things, but when \(h = 0\), we would get the relation \(X \otimes Y - Y \otimes X = 0\), which would leave us with things being commutative, i.e. we'd end up with the symmetric algebra \(Sym(\mf g)\) instead. So we can think of \(U(\mf g)\) as a deformed version of \(S(\mf g)\), which is not exactly commutative but where the noncommutativity is tightly controlled.
So let's consider a Lie group \(G\) and its Lie algebra \(\mf g\). Since we'll have several identities floating around, we'll write the identity element of \(G\) as \(e\), whereas identity matrices will be denoted \(I\).
For a vector space \(V\), \(End(V)\) has a Lie bracket, where for matrices \(S\) and \(T\) we have \([S, T] = ST - TS\). A Lie algebra action of \(\mf g\) on a vector space \(V\) is a map \(\rho: \mf g \rar End(V)\) such that the bracket on \(\mf g\) becomes the bracket on \(End(V)\). In other words,
$$\rho([X,Y]) = \rho(X)\rho(Y) - \rho(Y)\rho(X).$$Given an action of \(G\) on a vector space \(V\) with map \(\rho\), we can make a Lie algebra action by saying that for \(X \in \mf g\),
$$\rho(X) = \frac{\rho(e + sX) - \rho(e)}{s}$$This looks like a derivative, and those less comfortable with infinitesimals may replace it with such.
Note that this is not an algebra action, at least not if we use the Lie bracket as the multiplication, because the bracket in the Lie algebra does not become composition in \(End(V)\).
The comultiplication on \(G\) sends \(g\) to \(g \otimes g\). In particular, the identity element \(e\) gets sent to \(e \otimes e\). We view the Lie algebra as elements \(X\) such that \(e + sX\) is in \(G\) for infinitesimal \(s\). So
$$\Delta(e + sX) = (e + sX)\otimes (e + sX) = e \otimes e + sX \otimes e + e \otimes sX$$ where we use the fact that \(s^2 = 0\).
The point of infinitesimals is that they make everything look linear, so we have the idea that
$$\Delta(e + sX) = \Delta(e) + \Delta(sX) = \Delta(e) + s\Delta(X)$$Since as mentioned, \(\Delta(e) = e \otimes e\), we get that \(\Delta(X) = X\otimes e + e \otimes X\). This comultiplication is cocommutative and coassociative.
Thus we get that for two representations \((V,\rho)\) and \((W, \sigma)\), \(\mf g\) acts on \(V\otimes W\) via
$$(\rho \hat \otimes \sigma)(X)(v \otimes w) = \rho(X)v \otimes w + v \otimes \sigma(X)w$$This should look somewhat product-rule-y, in keeping with the understanding of Lie algebras as derivatives of Lie groups.
Although the comultiplication sends \(G\) to \(G\otimes G\), we don't get that the comultiplication sends \(\mf g\) to \(\mf g \otimes \mf g\). Instead it gets sent to \((\mb ke \oplus \mf g) \otimes (\mb ke \oplus \mf g)\). Most of the time we'll leave off the \(e\).
We can make \(\mb k \oplus \mf g\) into a counital coalgebra by defining \(\epsilon(X) = 0\) for all \(X \in \mf g\) and \(\epsilon(c) = c\) for \(c \in \mb k\).
What if we want an algebra that contains the Lie algebra \(\mf g\) where Lie algebra actions of \(\mf g\) become algebra actions? We call such an object the universal enveloping algebra of \(\mf g\), denoted \(U(\mf g)\), and we create it as a quotient.
We start with the tensor algebra \(\bigotimes^* \mf g\), where we're viewing \(\mf g\) as a vector space. Now we impose the Lie algebra structure by looking at the ideal \(I = \left\langle X \otimes Y - Y \otimes X - [X, Y]\right\rangle\) where \([, ]\) is the Lie bracket, and forming the quotient \(U(\mf g) = \bigotimes^* \mf g/I\).
Recall that \(\bigotimes^* V\) has a comultiplication, given by sending \(v \in V\) to \(v \otimes 1 + 1 \otimes v\), and extending by multiplication. It also has a counit, sending \(v\) to \(0\), and an antipode that sends \(v\) to \(-v\). So \(\bigotimes^* V\) is a Hopf algebra. If \(V = \mf g\), then the ideal \(I\) defined above is both an ideal and a coideal and is closed under antipode, so the quotient \(U(\mf g)\) is not only an algebra, but a Hopf algebra.
Given a universal enveloping algebra, we can recover the Lie algebra that it is built from by looking for the elements \(X\) such that \(\Delta(X) = X \otimes 1 + 1 \otimes X\), which we call "primitive" elements. Note that \(\Delta\) respects the Lie bracket, in that
$$\Delta([X, Y]) = [X, Y] \otimes 1 + 1 \otimes [X, Y] = \Delta(X)\Delta(Y) - \Delta(Y)\Delta(X)$$Thus the primitive elements in a Hopf algebra do form a Lie algebra.
For our universal enveloping algebra, we can take a representation of \(\mf g\) and extend it to a representation of \(U(\mf g)\) as follows: any non-scalar element in \(U(\mf g)\) can be written as a sum of products of elements of \(\mf g\), so we focus on such products, writing them as \(X_1X_2\cdots X_k\). Given a representation \((V,\rho)\) of \(\mf g\), we apply \(\rho\) to \(X_1X_2\cdots X_k\) in the obvious fashion, as \(\rho(X_1)\circ \rho(X_2) \circ \cdots \circ \rho(X_k)\). Scalars map also in the obvious fashion, with \(c \in \mb k \subset U(\mf g)\) going to \(cI\in End(V)\). Since \(\rho\) respects the Lie bracket, we get a well-defined algebra representation of \(U(\mf g)\). So now we can use algebra action theorems and techniques to talk about Lie algebra actions.
If we instead look at the ideal \(I_h = \left\langle X \otimes Y - Y\otimes X - h[X,Y]\right\rangle\) for \(h \in \mb k\), we get a slightly different algebra sometimes denoted by \(U_h(\mf g)\). For \(h\) nonzero this is isomorphic to \(U(\mf g)\) just by rescaling things, but when \(h = 0\), we would get the relation \(X \otimes Y - Y \otimes X = 0\), which would leave us with things being commutative, i.e. we'd end up with the symmetric algebra \(Sym(\mf g)\) instead. So we can think of \(U(\mf g)\) as a deformed version of \(S(\mf g)\), which is not exactly commutative but where the noncommutativity is tightly controlled.
Sunday, February 7, 2016
Quotients
\(\newcommand{\mb}{\mathbb}\)Let's talk about dividing sets by other sets.
You've probably heard of modular arithmetic. The usual first explanation is either in terms of clocks or remainders, or sometimes both. You say "if I multiply 5 by 9 modulo 13, I get 6" because 5 * 9 = 45, and then we subtract 13s and end up with 6.
Here's another way to look at it: consider a map\(f\) from \(\mb Z\) to a set \(T\) that has 13 elements in it, where \(f\) has the property that if \(a\) and \(b\) differ by an integer multiple of \(13\), then \(f(a) = f(b)\), and if they don't differ by an integer multiple of \(13\) exactly, then \(f\) sends them to different places. So \(f(1) = f(14)\) and \(f(45) = f(6)\), but \(f(1)\) and \(f(45)\) are different. Note that each element of \(T\) has to get hit by something, i.e. \(f\) is a surjection.
Now we give \(T\) some structure. We'll define an addition on \(T\) via addition on \(\mb Z\). In other words, we'll say that \(f(a) + f(b) = f(a + b)\) so that \(f\) is a homomorphism of groups with addition. Since \(f\) hits everything in \(T\), this defines addition for everything in \(T\). The question is whether this is well-defined; if we have \(A \in T\), there are several possible values \(a\) in \(\mb Z\) such that \(f(a) = A\), and similarly for \(B\) in \(T\) there are several values \(b\) such that \(f(b) = B\), and we want to make sure that \(A + B\) gives the same value no matter which \(a\) and \(b\) we pick. The conditions on \(f\) ensure this works in this case.
We can now consider general surjections \(f\) from \(\mb Z\) to other sets \(R\). If we want \(f\) to be a homomorphism, we need to have an addition structure on \(R\), with an identity and additive inverses, but we also need conditions on \(f\). In particular, let's look at \(f(0)\); this has to be the additive identity of \(R\). We call the set of things that get mapped to \(f(0)\) the kernel of \(f\), \(\ker(f)\); for the case above, this kernel is \(13\mb Z = \{13 k | k \in \mb Z\}\). In general, if \(f\) is going to be a homomorphism from \(\mb Z\) to another group , we need that \(\ker(f)\) is a subgroup of \(\mb Z\).
Let's suppose that we're looking at a general group \(G\) and a surjective homomorphism \(f\) from \(G\) to \(H\), and look at \(\ker(f)\), everything that gets sent to the identity \(e\) in \(H\). \(\ker(f)\) needs to be a group, because the identity composed with the identity is the identity.
But it needs to be more than just a subgroup of \(G\).
Consider \(g\) and \(h\) in \(G\) where \(h \in \ker(f)\). \(f(gh) = f(g)f(h) = f(g)\), since \(f\) is a homomorphism and \(f(h)\) is the identity. So everything of the form \(gh\) for \(h \in \ker(f)\) gets sent to the same place. Moreover, suppose that \(f(g') = f(g)\) for some \(g'\). Then we get that \(f(g^{-1}g') = f(g^{-1}f(g') = f(g^{-1})f(g') = f(g)^{-1}f(g') = f(g)^{-1}f(g) = e\). So \(g^{-1}g'\) is in \(\ker(f)\). Multiplying it by \(g\) gives that \(g'\) is of the form \(gh\) for some \(h \in \ker(f)\). We call the set \(gker(f) = \{gh | h \in \ker(f)\}\) a left-coset of \(\ker(f)\) in \(G\).
A similar argument tells us that \(g'\) is of the form \(h'g\) for some \(h' \in \ker(f)\), i.e. \(g'\) must be in the right-coset \(\ker(f)g\). So in order for \(f\) to be a homomorphism, we need that the left and right cosets \(g\ker(f)\) and \(ker(f)g\}\) are the same for all \(g\) in \(G\). We thus say that \(\ker(f)\) is normal in \(G\).
Conversely, if we have a normal subgroup \(N\), i.e. if for each \(g \in G\) the sets \(gN = \{gn | n \in N\}\) and \(Ng = \{ng | n \in N\}\) contain the same elements, then we can make a surjective homomorphism \(f\) from \(G\) to a group denoted \(G/N\) whose elements are cosets \(gN\), with multiplication in \(G/N\) given by \(gNhN = ghN\). We call \(G/N\) the quotient group of \(G\) by \(N\).
In the case of modular arithmetic above, if we treat \(\mb Z\) as a group whose composition operation is addition, then we get that all subgroups of \(\mb Z\) are normal. Indeed, for any group whose composition operation is commutative, all subgroups are normal. So for instance, given a vector space, \(V\) and a subspace \(W\), we can form \(V/W\) whose elements are cosets of the form \(v + W = \{v + w | w \in W\}\). Note that in this case we can also scalar multiply, since if \(w \in W\) then \(cw \in W\) for any scalar \(c\).
For groups that aren't commutative, the situation is more complicated. For example, consider the group \(S_3\) of permutations of 3 elements \(\{a,b,c\}\). The subgroup that contains just the identity and the permutation that swaps \(a\) and \(b\) is not normal. The subgroup \(C\) that contains the identity and the two ways to cycle all 3 elements is normal, and the quotient \(S/C\) is a group with two elements in it.
What about multiplication in quotients of \(\mb Z\)? Well, that always works so it's boring, so let's generalize until it sometimes breaks.
Consider a general associative ring \(R\), i.e. a set with an addition, additive identity, additive inverses, and a multiplication. Think matrices, for instance, or algebras.
We have a surjection \(f\) to another ring and we want to make this a homomorphism; what can we say about \(\ker(f)\)?
\(R\) is a group if we only look at addition, so \(\ker(f)\) must be a subgroup; the addition is commutative, so \(\ker(f)\) is automatically normal for addition. Our elements of \(R/\ker(f)\) are cosets of the form \(r + \ker(f)\).
Recalling that everything in \(\ker(f)\) is equivalent to \(0\) according to our map \(f\), we should get that \(\ker(f)\) times itself should be equivalent to \(0*0 = 0\), so we want that \(\ker(f)^2 = \{ab | a, b \in \ker(f)\}\) should be contained in \(\ker(f)\). So \(\ker(f)\) has to be a ring. But just as we needed more than just a subgroup, we need more than just a subring.
In our quotient object, we need that when we multiply something by anything equivalent to 0, the product is equivalent to 0. In other words, we need that \(a\ker(f) = \{ab | b \in \ker(f)\}\) needs to be contained in \(\ker(f)\). Similarly, we need that \(\ker(f)b = \{ab | a \in \ker(f)\}\) also needs to be contained in \(\ker(f)\). We say that \(\ker(f)\) is therefore a two-sided ideal, which I'll just refer to as an ideal.
Being an ideal in a ring is the equivalent of being a normal subgroup for groups: if we have a ring \(R\) and an ideal \(I\), then we can form the quotient ring \(R/I\) whose elements are of the form \(r + I\) and we have a natural surjective homomorphism \(f\) from \(R\) to \(R/I\) whose kernel is \(I\).
This is how we can think about modular arithmetic: for an integer \(k\), \(k\mb Z\) is an ideal in \(\mb Z\), so \(\mb Z/k\mb Z\) is a ring with addition, subtraction, and multiplication modulo \(k\).
We can think of this in terms of actions: A group \(G\) acts on itself via what is called the adjoint action or conjugation action: \(\rho(g)\) is the map that sends \(h\) to \(ghg^{-1}\). \(H\) being normal is then equivalent to saying that the set \(H\) is fixed as a whole by the adjoint action, although the individual elements of \(H\) might get shuffled around. Note that for commutative groups, the adjoint action is trivial, so all subgroups are fixed under it.
Similarly, a ring \(R\) acts on itself via either the left regular action or right regular action, also known as multiplication: \(lm(a)\) is the map that sends \(b\) to \(ab\), \(rm(b)\) is the map that sends \(a\) to \(ab\). \(I\) being an ideal is then equivalent to saying that \(I\) is invariant under both the left and right regular actions.
This invariance then allows for making quotients.
Given a bunch of elements \(\{x_1,\ldots,x_k\}\) in \(R\), we often write \(\left\langle x_1,\ldots,x_k\right\rangle\) for the smallest ideal in \(R\) that contains all of those elements. For groups, the same notation means the smallest group that contains all of those elements, although this makes no claims to normality. We say that \(\left\langle x_1,\ldots,x_k\right\rangle\) is generated by the elements \(x_1,\ldots,x_k\).
This gives a good way to talk about constructions of groups or algebras: we start with a thing \(F\) with no structures or relations other than the ones guaranteed by the axioms, called a "free" thing. So for instance a set that obeys the group axioms but has no other relations is called a free group. \(\mb Z\) with addition is a free group; everything in \(\mb Z\) has to be there because the group axiom demands that we can form any power of an element, but we don't get "a = b" except in the cases demanded by the associativity and inverse laws.
Then if we want to impose a relation \(A = B\) where \(A\) and \(B\) are expressions, we form the invariant subthing \(I\) generated by the difference between \(A\) and \(B\), and the quotient \(F/I\) thus can't tell the difference between \(A\) and \(B\).
For instance, suppose we start with a vector space \(V\) and form the tensor algebra \(\bigotimes^* V\). This is an associative, unital algebra, but has no other relations between elements; there are no equations that aren't just due to linearity in \(V\), so \(\bigotimes^* V\) is the free associative algebra on \(V\).
Now suppose we want to make a commutative algebra, i.e. \(a\otimes b = b\otimes a\). We start with \(\bigotimes^* V\) and consider the ideal \(I = \left\langle a\otimes b - b\otimes a | a, b \in V\right\rangle\). Then \(S(V) = \bigotimes^* V/I\) is commutative, because we've quotiented out the difference between \(a\otimes b\) and \(b \otimes a\); \(S(V)\) is often called the symmetric algebra on \(V\).
Finally, because of course we should talk about it, we can talk about quotients of coalgebras. Here we define a coideal \(I\) in a coalgebra \(C\) to be a subspace of \(C\) as a vector space with the property that
$$\Delta(I) \subset I \otimes C + C\otimes I$$ which is the dualized version of being an ideal. This is more coaction-y than action-y, using the left and right coregular coactions. Anyway, in this case we have that \(C/I\) is a coalgebra. We'll see an example of a quotient coalgebra in a bit.
We often say that a thing with no proper quotients is simple or irreducible. Such objects are like the atoms of whatever we're studying, because we can't break them into smaller pieces. And just as studying atoms is important for chemistry, studying the simple things is important for the study of things. Also just as there are many ways to put a given set of atoms together to make a molecule, there are often many ways to put a given set of simple things together to make a larger thing.
You've probably heard of modular arithmetic. The usual first explanation is either in terms of clocks or remainders, or sometimes both. You say "if I multiply 5 by 9 modulo 13, I get 6" because 5 * 9 = 45, and then we subtract 13s and end up with 6.
Here's another way to look at it: consider a map\(f\) from \(\mb Z\) to a set \(T\) that has 13 elements in it, where \(f\) has the property that if \(a\) and \(b\) differ by an integer multiple of \(13\), then \(f(a) = f(b)\), and if they don't differ by an integer multiple of \(13\) exactly, then \(f\) sends them to different places. So \(f(1) = f(14)\) and \(f(45) = f(6)\), but \(f(1)\) and \(f(45)\) are different. Note that each element of \(T\) has to get hit by something, i.e. \(f\) is a surjection.
Now we give \(T\) some structure. We'll define an addition on \(T\) via addition on \(\mb Z\). In other words, we'll say that \(f(a) + f(b) = f(a + b)\) so that \(f\) is a homomorphism of groups with addition. Since \(f\) hits everything in \(T\), this defines addition for everything in \(T\). The question is whether this is well-defined; if we have \(A \in T\), there are several possible values \(a\) in \(\mb Z\) such that \(f(a) = A\), and similarly for \(B\) in \(T\) there are several values \(b\) such that \(f(b) = B\), and we want to make sure that \(A + B\) gives the same value no matter which \(a\) and \(b\) we pick. The conditions on \(f\) ensure this works in this case.
We can now consider general surjections \(f\) from \(\mb Z\) to other sets \(R\). If we want \(f\) to be a homomorphism, we need to have an addition structure on \(R\), with an identity and additive inverses, but we also need conditions on \(f\). In particular, let's look at \(f(0)\); this has to be the additive identity of \(R\). We call the set of things that get mapped to \(f(0)\) the kernel of \(f\), \(\ker(f)\); for the case above, this kernel is \(13\mb Z = \{13 k | k \in \mb Z\}\). In general, if \(f\) is going to be a homomorphism from \(\mb Z\) to another group , we need that \(\ker(f)\) is a subgroup of \(\mb Z\).
Let's suppose that we're looking at a general group \(G\) and a surjective homomorphism \(f\) from \(G\) to \(H\), and look at \(\ker(f)\), everything that gets sent to the identity \(e\) in \(H\). \(\ker(f)\) needs to be a group, because the identity composed with the identity is the identity.
But it needs to be more than just a subgroup of \(G\).
Consider \(g\) and \(h\) in \(G\) where \(h \in \ker(f)\). \(f(gh) = f(g)f(h) = f(g)\), since \(f\) is a homomorphism and \(f(h)\) is the identity. So everything of the form \(gh\) for \(h \in \ker(f)\) gets sent to the same place. Moreover, suppose that \(f(g') = f(g)\) for some \(g'\). Then we get that \(f(g^{-1}g') = f(g^{-1}f(g') = f(g^{-1})f(g') = f(g)^{-1}f(g') = f(g)^{-1}f(g) = e\). So \(g^{-1}g'\) is in \(\ker(f)\). Multiplying it by \(g\) gives that \(g'\) is of the form \(gh\) for some \(h \in \ker(f)\). We call the set \(gker(f) = \{gh | h \in \ker(f)\}\) a left-coset of \(\ker(f)\) in \(G\).
A similar argument tells us that \(g'\) is of the form \(h'g\) for some \(h' \in \ker(f)\), i.e. \(g'\) must be in the right-coset \(\ker(f)g\). So in order for \(f\) to be a homomorphism, we need that the left and right cosets \(g\ker(f)\) and \(ker(f)g\}\) are the same for all \(g\) in \(G\). We thus say that \(\ker(f)\) is normal in \(G\).
Conversely, if we have a normal subgroup \(N\), i.e. if for each \(g \in G\) the sets \(gN = \{gn | n \in N\}\) and \(Ng = \{ng | n \in N\}\) contain the same elements, then we can make a surjective homomorphism \(f\) from \(G\) to a group denoted \(G/N\) whose elements are cosets \(gN\), with multiplication in \(G/N\) given by \(gNhN = ghN\). We call \(G/N\) the quotient group of \(G\) by \(N\).
In the case of modular arithmetic above, if we treat \(\mb Z\) as a group whose composition operation is addition, then we get that all subgroups of \(\mb Z\) are normal. Indeed, for any group whose composition operation is commutative, all subgroups are normal. So for instance, given a vector space, \(V\) and a subspace \(W\), we can form \(V/W\) whose elements are cosets of the form \(v + W = \{v + w | w \in W\}\). Note that in this case we can also scalar multiply, since if \(w \in W\) then \(cw \in W\) for any scalar \(c\).
For groups that aren't commutative, the situation is more complicated. For example, consider the group \(S_3\) of permutations of 3 elements \(\{a,b,c\}\). The subgroup that contains just the identity and the permutation that swaps \(a\) and \(b\) is not normal. The subgroup \(C\) that contains the identity and the two ways to cycle all 3 elements is normal, and the quotient \(S/C\) is a group with two elements in it.
What about multiplication in quotients of \(\mb Z\)? Well, that always works so it's boring, so let's generalize until it sometimes breaks.
Consider a general associative ring \(R\), i.e. a set with an addition, additive identity, additive inverses, and a multiplication. Think matrices, for instance, or algebras.
We have a surjection \(f\) to another ring and we want to make this a homomorphism; what can we say about \(\ker(f)\)?
\(R\) is a group if we only look at addition, so \(\ker(f)\) must be a subgroup; the addition is commutative, so \(\ker(f)\) is automatically normal for addition. Our elements of \(R/\ker(f)\) are cosets of the form \(r + \ker(f)\).
Recalling that everything in \(\ker(f)\) is equivalent to \(0\) according to our map \(f\), we should get that \(\ker(f)\) times itself should be equivalent to \(0*0 = 0\), so we want that \(\ker(f)^2 = \{ab | a, b \in \ker(f)\}\) should be contained in \(\ker(f)\). So \(\ker(f)\) has to be a ring. But just as we needed more than just a subgroup, we need more than just a subring.
In our quotient object, we need that when we multiply something by anything equivalent to 0, the product is equivalent to 0. In other words, we need that \(a\ker(f) = \{ab | b \in \ker(f)\}\) needs to be contained in \(\ker(f)\). Similarly, we need that \(\ker(f)b = \{ab | a \in \ker(f)\}\) also needs to be contained in \(\ker(f)\). We say that \(\ker(f)\) is therefore a two-sided ideal, which I'll just refer to as an ideal.
Being an ideal in a ring is the equivalent of being a normal subgroup for groups: if we have a ring \(R\) and an ideal \(I\), then we can form the quotient ring \(R/I\) whose elements are of the form \(r + I\) and we have a natural surjective homomorphism \(f\) from \(R\) to \(R/I\) whose kernel is \(I\).
This is how we can think about modular arithmetic: for an integer \(k\), \(k\mb Z\) is an ideal in \(\mb Z\), so \(\mb Z/k\mb Z\) is a ring with addition, subtraction, and multiplication modulo \(k\).
We can think of this in terms of actions: A group \(G\) acts on itself via what is called the adjoint action or conjugation action: \(\rho(g)\) is the map that sends \(h\) to \(ghg^{-1}\). \(H\) being normal is then equivalent to saying that the set \(H\) is fixed as a whole by the adjoint action, although the individual elements of \(H\) might get shuffled around. Note that for commutative groups, the adjoint action is trivial, so all subgroups are fixed under it.
Similarly, a ring \(R\) acts on itself via either the left regular action or right regular action, also known as multiplication: \(lm(a)\) is the map that sends \(b\) to \(ab\), \(rm(b)\) is the map that sends \(a\) to \(ab\). \(I\) being an ideal is then equivalent to saying that \(I\) is invariant under both the left and right regular actions.
This invariance then allows for making quotients.
Given a bunch of elements \(\{x_1,\ldots,x_k\}\) in \(R\), we often write \(\left\langle x_1,\ldots,x_k\right\rangle\) for the smallest ideal in \(R\) that contains all of those elements. For groups, the same notation means the smallest group that contains all of those elements, although this makes no claims to normality. We say that \(\left\langle x_1,\ldots,x_k\right\rangle\) is generated by the elements \(x_1,\ldots,x_k\).
Then if we want to impose a relation \(A = B\) where \(A\) and \(B\) are expressions, we form the invariant subthing \(I\) generated by the difference between \(A\) and \(B\), and the quotient \(F/I\) thus can't tell the difference between \(A\) and \(B\).
For instance, suppose we start with a vector space \(V\) and form the tensor algebra \(\bigotimes^* V\). This is an associative, unital algebra, but has no other relations between elements; there are no equations that aren't just due to linearity in \(V\), so \(\bigotimes^* V\) is the free associative algebra on \(V\).
Now suppose we want to make a commutative algebra, i.e. \(a\otimes b = b\otimes a\). We start with \(\bigotimes^* V\) and consider the ideal \(I = \left\langle a\otimes b - b\otimes a | a, b \in V\right\rangle\). Then \(S(V) = \bigotimes^* V/I\) is commutative, because we've quotiented out the difference between \(a\otimes b\) and \(b \otimes a\); \(S(V)\) is often called the symmetric algebra on \(V\).
Finally, because of course we should talk about it, we can talk about quotients of coalgebras. Here we define a coideal \(I\) in a coalgebra \(C\) to be a subspace of \(C\) as a vector space with the property that
$$\Delta(I) \subset I \otimes C + C\otimes I$$ which is the dualized version of being an ideal. This is more coaction-y than action-y, using the left and right coregular coactions. Anyway, in this case we have that \(C/I\) is a coalgebra. We'll see an example of a quotient coalgebra in a bit.
We often say that a thing with no proper quotients is simple or irreducible. Such objects are like the atoms of whatever we're studying, because we can't break them into smaller pieces. And just as studying atoms is important for chemistry, studying the simple things is important for the study of things. Also just as there are many ways to put a given set of atoms together to make a molecule, there are often many ways to put a given set of simple things together to make a larger thing.
Coactions
\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)Just as we can dualize things to turn an algebra into a coalgebra, we can dualize an action to get a coaction.
A left coaction of a set \(S\) on a space \(V\) is usually written as a map \(\beta: V \rar S \otimes V\), which we compare to an action of \(T\) on \(V\) which is written \(T \otimes V \rar V\). If \(V\) is finite-dimensional, then we can turn the coaction map into a map \(End(V^\vee) \rar S\). We say that \(V\) is an \(S\)-comodule with coaction map \(\beta\).
If \(S\) can be dualized, for instance if it is a finite-dimensional vector space, then we can interpret a coaction of \(S\) as an action of \(S^\vee\). In particular, if the coaction of \(S\) is the map \(\beta: V \rar S\otimes V\) then the induced action of \(S^\vee\) is given by \(\rho(t \otimes v) = (t \otimes id)(\beta(v))\). Dually, an action of \(S^\vee\) gives a coaction of \(S\), by setting \(\beta(v) = e_i\otimes \rho(e^i \otimes v)\) where \(e_i\) is a basis of \(S\) and \(e^i\) is the corresponding dual basis of \(S^\vee\).
We can also talk about algebra coactions and coalgebra coactions. The rules for an algebra coaction are dual to the rules for a coalgebra action. Namely, if we have an algebra \(A\) and two coactions of \(A\) labeled \((V, \beta)\) and \((W, \gamma)\), we can make a algebra coaction on \(V\otimes W\) by setting
$$\beta \hat \otimes \gamma = (m \otimes id \otimes id) \circ (id \otimes \sigma_{VA}\otimes id) \circ(\beta \otimes \gamma)$$In other words, if \(\beta(v) = a_i \otimes v^i\) and \(\gamma(w) = a_j \otimes w^j\), then we get that
$$(\beta \hat \otimes \gamma)(v \otimes w) = a_ia_j \otimes v^i \otimes w^j$$
The rules for a coalgebra coaction are dual to the rules for an algebra action. In other words,
$$(\Delta \otimes id) \circ \beta = (id \otimes \beta) \circ \beta$$
Just to establish it, we say that the trivial coaction of a unital thing on \(V\) sends \(v \in V\) to \(1 \otimes v\), just as the trivial action of a counital thing was the map \(s\otimes v\mapsto \epsilon(s)v\).
Coactions are often useful when the coacting set doesn't have a nice dual, perhaps due to being infinite-dimensional. Sometimes the coacting set is simply nicer to work with. For instance, instead of dealing with a group algebra, which can be badly behaved for infinite groups like Lie groups, we can instead look at the representation ring \(Rep(\mb kG)\) of \(mb kG\), i.e. the algebra generated by the set of functions \(\pi_i^j\) where \(\pi\) is the representation map for some finite-dimensional representation and \(\pi_i^j(\hat g)\) is the \((i,j)\) entry of \(\pi(\hat g)\).
Notably, if \(\rho\) is the representation map of some faithful finite-dimensional representation, then \(Rep(\mb kG)\) is generated by the \(\rho_i^j\). We'll talk about \(Rep\) a bit more later.
And just to show again that the notion of a coaction is really dual to actions, here's a bunch of diagrams; the solid line is the coacting set, the dashed is the set being coacted upon. The coaction itself is given by a black box. If you compare with the action diagrams, you'll see that they are in fact dual.
A left coaction of a set \(S\) on a space \(V\) is usually written as a map \(\beta: V \rar S \otimes V\), which we compare to an action of \(T\) on \(V\) which is written \(T \otimes V \rar V\). If \(V\) is finite-dimensional, then we can turn the coaction map into a map \(End(V^\vee) \rar S\). We say that \(V\) is an \(S\)-comodule with coaction map \(\beta\).
If \(S\) can be dualized, for instance if it is a finite-dimensional vector space, then we can interpret a coaction of \(S\) as an action of \(S^\vee\). In particular, if the coaction of \(S\) is the map \(\beta: V \rar S\otimes V\) then the induced action of \(S^\vee\) is given by \(\rho(t \otimes v) = (t \otimes id)(\beta(v))\). Dually, an action of \(S^\vee\) gives a coaction of \(S\), by setting \(\beta(v) = e_i\otimes \rho(e^i \otimes v)\) where \(e_i\) is a basis of \(S\) and \(e^i\) is the corresponding dual basis of \(S^\vee\).
We can also talk about algebra coactions and coalgebra coactions. The rules for an algebra coaction are dual to the rules for a coalgebra action. Namely, if we have an algebra \(A\) and two coactions of \(A\) labeled \((V, \beta)\) and \((W, \gamma)\), we can make a algebra coaction on \(V\otimes W\) by setting
$$\beta \hat \otimes \gamma = (m \otimes id \otimes id) \circ (id \otimes \sigma_{VA}\otimes id) \circ(\beta \otimes \gamma)$$In other words, if \(\beta(v) = a_i \otimes v^i\) and \(\gamma(w) = a_j \otimes w^j\), then we get that
$$(\beta \hat \otimes \gamma)(v \otimes w) = a_ia_j \otimes v^i \otimes w^j$$
The rules for a coalgebra coaction are dual to the rules for an algebra action. In other words,
$$(\Delta \otimes id) \circ \beta = (id \otimes \beta) \circ \beta$$
Just to establish it, we say that the trivial coaction of a unital thing on \(V\) sends \(v \in V\) to \(1 \otimes v\), just as the trivial action of a counital thing was the map \(s\otimes v\mapsto \epsilon(s)v\).
Coactions are often useful when the coacting set doesn't have a nice dual, perhaps due to being infinite-dimensional. Sometimes the coacting set is simply nicer to work with. For instance, instead of dealing with a group algebra, which can be badly behaved for infinite groups like Lie groups, we can instead look at the representation ring \(Rep(\mb kG)\) of \(mb kG\), i.e. the algebra generated by the set of functions \(\pi_i^j\) where \(\pi\) is the representation map for some finite-dimensional representation and \(\pi_i^j(\hat g)\) is the \((i,j)\) entry of \(\pi(\hat g)\).
Notably, if \(\rho\) is the representation map of some faithful finite-dimensional representation, then \(Rep(\mb kG)\) is generated by the \(\rho_i^j\). We'll talk about \(Rep\) a bit more later.
And just to show again that the notion of a coaction is really dual to actions, here's a bunch of diagrams; the solid line is the coacting set, the dashed is the set being coacted upon. The coaction itself is given by a black box. If you compare with the action diagrams, you'll see that they are in fact dual.
Coalgebra actions
\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)I know that most everyone who has already learned group representation theory is probably thinking "why are you talking about coalgebras now? Just list a bunch of representations of groups using the usual setup". But where's the fun in that?
As we saw, an algebra action of an algebra \(A\) on a vector space \(V\) turned the multiplication in \(A\) into composition in \(End(V)\), So let's see what that says about coalgebras.
We take \(\Delta\) in a coalgebra \(C\). Does \(End(V)\) have a comultiplication (cocomposition?) that we can map \(\Delta\) to? If \(V\) is infinite-dimensional, not really.
Instead we get a multiplication operation on representations.
Take two representations \(\rho, V\) and \(\sigma, W\). Then we can make the representation \((\rho \hat \otimes \sigma, V\otimes W)\), where for \(f \in C\),
$$(\rho \hat \otimes \sigma)(f) = (\rho \otimes \sigma)\Delta(f)$$In other words, we take the comultiplication of \(f\), apply \(\rho\) to the first factor and \(\sigma\) to the second factor, getting an element of \(End(V \otimes W)\).
So the coalgebra structure, rather than restricting actions the way an algebra structure does, instead allows for the easy creation of new ones via the tensor product. Diagrammatically:
So for instance, if we have a matrix group \(G\), there's a vector space \(V\) on which it acts naturally via the defining representation \(\rho\), and \(\mb kG\) also acts on \(V\) via \(\rho\). Since \(\mb kG\) has a natural comultiplication that sends \(\hat g\) to \(\hat g \otimes \hat g\), we get that \(\mb kG\) acts on \(V\otimes V\) by
$$(\rho \hat \otimes \rho)(g)(u\otimes g) = \rho(g)u\otimes \rho(g)v$$And similarly it acts on \(V\otimes V\otimes V\) etc.
This allows us to have some fun with putting structures on \(V\).
Suppose that we have an algebra \(A\) and a coalgebra \(C\). We say that a coalgebra action \(\rho\) of \(C\) on \(A\) is compatible with the algebra structure of \(A\) if it plays well with the algebra structure of \(A\). In other words,
$$\rho(f)(m_A(a\otimes b)) = m_A((\rho \hat \otimes \rho)(f) \textbf{ }(a \otimes b))$$where \(m_A\) is the multiplication on \(A\).
We then say that \(A\) is then a \(C\)-module algebra.
So for a group \(G\), for instance, we say that it acts on an algebra \(A\) if we have that \(\rho(g)(m_A(a\otimes b)) = m_A(\rho(g)a\otimes \rho(g)b)\). This allows us to talk about the automorphism group of an algebra: defining \(GL(A)\) to be the general linear group for \(A\) treated as a vector space, we look at the subgroup of all \(g\) such that the compatibility rule for acting on an algebra holds. We call the resulting group \(Aut(A)\).
We can also talk about a coalgebra acting on a coalgebra. Suppose that \(C\) acts on \(H\), both coalgebras. Then we ask that the comultiplications and the action map play well:
$$\Delta_H(\rho(f)h) = (\rho \hat \otimes \rho)(f) \Delta_H(h)$$where \(\Delta_H\) is the comultiplication on \(H\), and we say that \(H\) is a \(C\)-module coalgebra.
Again we can define \(Aut(H)\) for the coalgebra \(H\) as the subgroup of \(GL(H)\) for which that rule holds.
By all of this, we can define a coalgebra acting on a bialgebra by combining all of these rules. By combining the coalgebra action rules with the algebra action rules, we get the rules for the action of a bialgebra.
The diagrams for a coalgebra acting on an algebra and a coalgebra look like:
if you'll forgive using the same black dot to indicate comultiplication in both \(C\) and \(H\). The lower equation should look like the multiplication-comultiplication compatibility diagram for Hopf algebras. The upper equation doesn't have such a nice analogy in general.
What about Hopf algebras, in particular the antipode map? Well, we want any action on a Hopf algebra to preserve the antipode map:
$$\rho(g)(S_Vv) = S_V(\rho(g)v)$$which diagrammatically just means that you can move \(S\)s on the dashed lines past the white boxes.
As for a Hopf algebra acting on a vector space \(V\), having \(S_H\) allows for actions on the dual of \(V\); if \(S_H\) is invertible, we actually get two actions, one using \(S_H\) and one using \(S_H^{-1}\).
Let's take \((V,\rho)\) as a representation of \(H\), and let \((V^\vee, \rho^\vee)\) be the representation of \(H\) where for \(f \in V^\vee\), \(\rho^\vee(g)(f) = f \circ \rho(S_H(g))\). Similarly, let \((^\vee V, ^\vee\rho)\) have the action \(^\vee\rho(g)(f) = f \circ \rho(S_H^{-1}(g))\). You can check that both of these are actions. For the objects that we're familiar with, groups and Lie algebras and commutative rings, \(S_H = S_H^{-1}\) and so there's only one action on the dual space, but for more general Hopf algebras they don't have to be the same.
You might also consider higher powers of \(S_H\), but we're not going to.
As we saw, an algebra action of an algebra \(A\) on a vector space \(V\) turned the multiplication in \(A\) into composition in \(End(V)\), So let's see what that says about coalgebras.
We take \(\Delta\) in a coalgebra \(C\). Does \(End(V)\) have a comultiplication (cocomposition?) that we can map \(\Delta\) to? If \(V\) is infinite-dimensional, not really.
Instead we get a multiplication operation on representations.
Take two representations \(\rho, V\) and \(\sigma, W\). Then we can make the representation \((\rho \hat \otimes \sigma, V\otimes W)\), where for \(f \in C\),
$$(\rho \hat \otimes \sigma)(f) = (\rho \otimes \sigma)\Delta(f)$$In other words, we take the comultiplication of \(f\), apply \(\rho\) to the first factor and \(\sigma\) to the second factor, getting an element of \(End(V \otimes W)\).
So the coalgebra structure, rather than restricting actions the way an algebra structure does, instead allows for the easy creation of new ones via the tensor product. Diagrammatically:
$$(\rho \hat \otimes \rho)(g)(u\otimes g) = \rho(g)u\otimes \rho(g)v$$And similarly it acts on \(V\otimes V\otimes V\) etc.
This allows us to have some fun with putting structures on \(V\).
Suppose that we have an algebra \(A\) and a coalgebra \(C\). We say that a coalgebra action \(\rho\) of \(C\) on \(A\) is compatible with the algebra structure of \(A\) if it plays well with the algebra structure of \(A\). In other words,
$$\rho(f)(m_A(a\otimes b)) = m_A((\rho \hat \otimes \rho)(f) \textbf{ }(a \otimes b))$$where \(m_A\) is the multiplication on \(A\).
We then say that \(A\) is then a \(C\)-module algebra.
So for a group \(G\), for instance, we say that it acts on an algebra \(A\) if we have that \(\rho(g)(m_A(a\otimes b)) = m_A(\rho(g)a\otimes \rho(g)b)\). This allows us to talk about the automorphism group of an algebra: defining \(GL(A)\) to be the general linear group for \(A\) treated as a vector space, we look at the subgroup of all \(g\) such that the compatibility rule for acting on an algebra holds. We call the resulting group \(Aut(A)\).
We can also talk about a coalgebra acting on a coalgebra. Suppose that \(C\) acts on \(H\), both coalgebras. Then we ask that the comultiplications and the action map play well:
$$\Delta_H(\rho(f)h) = (\rho \hat \otimes \rho)(f) \Delta_H(h)$$where \(\Delta_H\) is the comultiplication on \(H\), and we say that \(H\) is a \(C\)-module coalgebra.
Again we can define \(Aut(H)\) for the coalgebra \(H\) as the subgroup of \(GL(H)\) for which that rule holds.
By all of this, we can define a coalgebra acting on a bialgebra by combining all of these rules. By combining the coalgebra action rules with the algebra action rules, we get the rules for the action of a bialgebra.
The diagrams for a coalgebra acting on an algebra and a coalgebra look like:
What about Hopf algebras, in particular the antipode map? Well, we want any action on a Hopf algebra to preserve the antipode map:
$$\rho(g)(S_Vv) = S_V(\rho(g)v)$$which diagrammatically just means that you can move \(S\)s on the dashed lines past the white boxes.
As for a Hopf algebra acting on a vector space \(V\), having \(S_H\) allows for actions on the dual of \(V\); if \(S_H\) is invertible, we actually get two actions, one using \(S_H\) and one using \(S_H^{-1}\).
Let's take \((V,\rho)\) as a representation of \(H\), and let \((V^\vee, \rho^\vee)\) be the representation of \(H\) where for \(f \in V^\vee\), \(\rho^\vee(g)(f) = f \circ \rho(S_H(g))\). Similarly, let \((^\vee V, ^\vee\rho)\) have the action \(^\vee\rho(g)(f) = f \circ \rho(S_H^{-1}(g))\). You can check that both of these are actions. For the objects that we're familiar with, groups and Lie algebras and commutative rings, \(S_H = S_H^{-1}\) and so there's only one action on the dual space, but for more general Hopf algebras they don't have to be the same.
You might also consider higher powers of \(S_H\), but we're not going to.
Algebra actions
\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)We started talking about groups in terms of what they do to things: rotating a cube or a peg, etc. Matrix groups have natural examples of doing stuff to vectors.
Let's formalize a little.
An action of a set \(S\) on a set \(T\) is a map \(\rho\) that assigns to each \(s \in S\) a map from \(T\) to itself, usually obeying some rules about the structures of \(S\) and \(T\). If we fix \(s \in S\), then the map \(t \mapsto \rho(s, t)\) is a map from \(T\) to \(T\). So for each \(s \in S\), \(\rho\) can be used to get an element of \(End(T)\), the set of maps from \(T\) to itself.
We say that an action is faithful if the map from \(S\) to \(End(T)\) is injective, i.e. if \(\rho(s) = \rho(s')\) then \(s = s'\).
In diagrams, we'd write the action map using a white square: the solid line takes in an element of \(S\), the dashed is for elements of \(T\).
We usually think of \(End(T)\) as respecting various structures that \(T\) may have. For instance, if \(T\) is a vector space, then \(End(T)\) is the set of linear maps from \(T\) to itself, which for finite-dimensional \(T\) manifests a square matrices. Hence we would like that \(\rho(s)\) is a linear transformation for all \(s\in S\).
In fact, we're going to nail that down right now: all of our actions are on vector spaces.
Given a vector space \(V\) and an action map \(\rho\) of \(S\) on \(V\), we often say that \((V, \rho)\) is a representation of \(S\). We call \(V\) an \(S\)-module. This actually is a bit of a mix of terminology; a representation theorist will usually say that \(V\) is a representation, or affords a representation, rather than saying that the pair \((V,\rho)\) is a representation. The term module is usually used in algebra and related disciplines like algebraic geometry and refers to slightly more general structures that don't necessarily have built-in scalar multiplication. So the distinction I'm making here isn't entirely standard convention, but I think it to be useful.
What if \(S\) has structure as well? If it's a structure that \(End(T)\) carries, then we usually want \(\rho\) to respect that structure. For instance, if \(S\) is a vector space, then we want \(\rho\) to be a linear map from \(S\) to \(End(T)\), which is also a vector space.
Given a group \(G\), we can talk about a group action or group representation \(\rho\) of \(G\) on a vector space \(V\) saying that for each \(g \in G\), \(\rho(g)\) is in \(End(V)\). But \(G\) is more than just a set, it's a group, so we need a few things.
First, we want \(\rho\) to send group composition to composition of maps, so we say that, as maps from \(V\) to itself,
$$\rho(gh) = \rho(g)\circ\rho(h)$$Also, because it makes sense to, we'll say that for the identity element \(e \in G\), we have
$$\rho(e) = id_V$$Because where else should the identity go but to the identity?
So a group representation is an action of \(G\) on \(V\) that obeys those two laws. Note that this means that \(\rho(g^{-1}) = \rho(g)^{-1}\) automatically. Furthermore, that means that \(\rho(g)\) is always invertible. Hence \(G\) ends up in the subset \(GL(V)\) contained in \(End(V)\).
What about an algebra? What about a representation \(\rho\) of an algebra \(A\) on \(V\)?
An algebra is a vector space, so we say that the map \(\rho: A\rar End(V)\) is linear. An algebra also has a multiplication map, and \(End(V)\) has a composition operation, so we say that \(\rho\) should send multiplication to composition, just as with group actions. In other words,
$$\rho(ab) = \rho(a)\circ\rho(b)$$If \(A\) is unital, then we add in that for \(c \in \mb k\),
$$\rho(\eta(c)) = cId_V$$Note that even if \(A\) is not associative, \(End(V)\) is. So \(\rho(a(bc))\) and \(\rho((ab)c)\) end up going to the same thing: \(\rho(a)\rho(b)\rho(c)\). On the other hand, \(End(V)\) isn't generally commutative, so we can still detect a difference between \(\rho(ab)\) and \(\rho(ba)\).
From a group representation \(\rho: G\rar End(V)\), we easily get an algebra representation of \(\mb kG\) just by sending \(\sum_{g \in G} a_g \hat g\) to \(\sum_{g \in G} a_g \rho(g)\).
Diagrammatically, we have
Note how similar these are to the associativity and unit diagrams for algebras, only now instead of just white circles we also have some white squares.
Examples!
The trivial representation is of \(G\) or \(\mb kG\) acting on \(\mb k\) with everything in \(G\), or correspondingly all elements of the form \(\hat g\), acting as the identity; for \(\mb kG\) we then extend by linearity. This is boring, but is important enough that we mention it.
Matrix groups are naturally subgroups of \(GL(V)\) for some \(V\) by definition, and so naturally have representations.
Given a representation \(\rho: G \rar End(V)\), there is also a representation \(\rho^\vee\) on the dual vector space \(V^\vee\). Suppose that \(f \in V^\vee\). How should \(g\) act on \(f\)? We want it to be compatible with an action of \(g\) on \(V\). The compatibility is that acting on both \(f\) and \(v\) does nothing:
$$(\rho^\vee(g)f)(\rho(g)v) = f(v)$$Note that \(f(v)\) is in \(\mb k\), so we're thinking that the combined action on \(V\) and \(V^\vee\) is the trivial action.
Anyway, this leads to saying that
$$(\rho^\vee(g)f)(v) = f(\rho(g)^{-1}v)$$Note that this an action, in that
$$\rho^\vee(g)(\rho^\vee(h)f) = \rho^\vee(gh)f$$Check this!
Basically this is just the change-of-basis law for dual spaces; if one space changes by \(\rho(g)\), the dual space changes by \(\rho(g)^{-1}\).
More examples to come when we have a bit more machinery.
Let's formalize a little.
An action of a set \(S\) on a set \(T\) is a map \(\rho\) that assigns to each \(s \in S\) a map from \(T\) to itself, usually obeying some rules about the structures of \(S\) and \(T\). If we fix \(s \in S\), then the map \(t \mapsto \rho(s, t)\) is a map from \(T\) to \(T\). So for each \(s \in S\), \(\rho\) can be used to get an element of \(End(T)\), the set of maps from \(T\) to itself.
We say that an action is faithful if the map from \(S\) to \(End(T)\) is injective, i.e. if \(\rho(s) = \rho(s')\) then \(s = s'\).
In diagrams, we'd write the action map using a white square: the solid line takes in an element of \(S\), the dashed is for elements of \(T\).
In fact, we're going to nail that down right now: all of our actions are on vector spaces.
Given a vector space \(V\) and an action map \(\rho\) of \(S\) on \(V\), we often say that \((V, \rho)\) is a representation of \(S\). We call \(V\) an \(S\)-module. This actually is a bit of a mix of terminology; a representation theorist will usually say that \(V\) is a representation, or affords a representation, rather than saying that the pair \((V,\rho)\) is a representation. The term module is usually used in algebra and related disciplines like algebraic geometry and refers to slightly more general structures that don't necessarily have built-in scalar multiplication. So the distinction I'm making here isn't entirely standard convention, but I think it to be useful.
What if \(S\) has structure as well? If it's a structure that \(End(T)\) carries, then we usually want \(\rho\) to respect that structure. For instance, if \(S\) is a vector space, then we want \(\rho\) to be a linear map from \(S\) to \(End(T)\), which is also a vector space.
Given a group \(G\), we can talk about a group action or group representation \(\rho\) of \(G\) on a vector space \(V\) saying that for each \(g \in G\), \(\rho(g)\) is in \(End(V)\). But \(G\) is more than just a set, it's a group, so we need a few things.
First, we want \(\rho\) to send group composition to composition of maps, so we say that, as maps from \(V\) to itself,
$$\rho(gh) = \rho(g)\circ\rho(h)$$Also, because it makes sense to, we'll say that for the identity element \(e \in G\), we have
$$\rho(e) = id_V$$Because where else should the identity go but to the identity?
So a group representation is an action of \(G\) on \(V\) that obeys those two laws. Note that this means that \(\rho(g^{-1}) = \rho(g)^{-1}\) automatically. Furthermore, that means that \(\rho(g)\) is always invertible. Hence \(G\) ends up in the subset \(GL(V)\) contained in \(End(V)\).
What about an algebra? What about a representation \(\rho\) of an algebra \(A\) on \(V\)?
An algebra is a vector space, so we say that the map \(\rho: A\rar End(V)\) is linear. An algebra also has a multiplication map, and \(End(V)\) has a composition operation, so we say that \(\rho\) should send multiplication to composition, just as with group actions. In other words,
$$\rho(ab) = \rho(a)\circ\rho(b)$$If \(A\) is unital, then we add in that for \(c \in \mb k\),
$$\rho(\eta(c)) = cId_V$$Note that even if \(A\) is not associative, \(End(V)\) is. So \(\rho(a(bc))\) and \(\rho((ab)c)\) end up going to the same thing: \(\rho(a)\rho(b)\rho(c)\). On the other hand, \(End(V)\) isn't generally commutative, so we can still detect a difference between \(\rho(ab)\) and \(\rho(ba)\).
From a group representation \(\rho: G\rar End(V)\), we easily get an algebra representation of \(\mb kG\) just by sending \(\sum_{g \in G} a_g \hat g\) to \(\sum_{g \in G} a_g \rho(g)\).
Diagrammatically, we have
Examples!
The trivial representation is of \(G\) or \(\mb kG\) acting on \(\mb k\) with everything in \(G\), or correspondingly all elements of the form \(\hat g\), acting as the identity; for \(\mb kG\) we then extend by linearity. This is boring, but is important enough that we mention it.
Matrix groups are naturally subgroups of \(GL(V)\) for some \(V\) by definition, and so naturally have representations.
Given a representation \(\rho: G \rar End(V)\), there is also a representation \(\rho^\vee\) on the dual vector space \(V^\vee\). Suppose that \(f \in V^\vee\). How should \(g\) act on \(f\)? We want it to be compatible with an action of \(g\) on \(V\). The compatibility is that acting on both \(f\) and \(v\) does nothing:
$$(\rho^\vee(g)f)(\rho(g)v) = f(v)$$Note that \(f(v)\) is in \(\mb k\), so we're thinking that the combined action on \(V\) and \(V^\vee\) is the trivial action.
Anyway, this leads to saying that
$$(\rho^\vee(g)f)(v) = f(\rho(g)^{-1}v)$$Note that this an action, in that
$$\rho^\vee(g)(\rho^\vee(h)f) = \rho^\vee(gh)f$$Check this!
Basically this is just the change-of-basis law for dual spaces; if one space changes by \(\rho(g)\), the dual space changes by \(\rho(g)^{-1}\).
More examples to come when we have a bit more machinery.
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