\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)We've mentioned an invariant for \(SO(3,\mb R)\) before, the object \(x\otimes x + y\otimes y + z\otimes z\). We're going to look at more invariants for group and Lie algebra actions.
Recall that we said that \(x\otimes x + y\otimes y + z\otimes z\) is invariant because for any \(g \in SO(3,\mb)\), \(g\) sends \(x \otimes x + y\otimes y + z\otimes z\) to itself, where \(SO(3, \mb R)\) acts on \(\mb R^3 \otimes \mb R^3\) using the action we get from the coalgebra structure of \(SO(3, \mb R)\).
We generalize this notion for other groups: for a representation \((V, \rho)\) of a group \(G\), we say that a vector \(v \in V\) is invariant under the action of \(G\) if for all \(g \in G\),
We generalize this notion for other groups: for a representation \((V, \rho)\) of a group \(G\), we say that a vector \(v \in V\) is invariant under the action of \(G\) if for all \(g \in G\),
$$\rho(g)v = v.$$
Given a representation \((V, \rho)\) of \(G\), we say that a tensor \(T\) is a tensor invariant of \(G\) if it is invariant as an element of \(V^{\otimes m} \otimes (V^\vee)^{\otimes k}\) with the action given by the coalgebra structure of \(G\).
Invariants of group actions provide a good way to understand the groups in question. Invariant tensors describe structures that are preserved by the group, giving a connection between symmetry and algebra and geometry.
We can generalize the notion of an invariant further to Hopf algebras: given a representation \((V, \rho)\) of a Hopf algebra \(H\), we say that a vector \(v \in V\) is invariant under the action of \(H\) if for all \(h \in H\),
$$\rho(h)v = \epsilon(h)v.$$
The appearance of \(\epsilon(h)v\) as opposed to just \(v\) allows for linearity. Similarly for the notion of an invariant tensor.
So for instance, for a Lie algebra \(\mf g\), we would say that a vector is invariant under the Lie algebra action if for all \(X \in \mf g\),
So for instance, for a Lie algebra \(\mf g\), we would say that a vector is invariant under the Lie algebra action if for all \(X \in \mf g\),
$$\rho(X)v = 0.$$
Let's look at some examples.
Every action leaves the identity matrix invariant, because otherwise things fail to be defined. If two things are equal, then the results after applying the action should also be equal. So \(I_i^j e_j \otimes e^i \in V\otimes V^\vee\) is an invariant for any representation \((V,\rho)\).
For \(SO(n)\), we have the example above of the tensor \(g = g_{ij}e^i \otimes e^j \in V^\vee \otimes V^\vee\), where \(V = \mb k^n\). In the standard basis, we would have \(g_{ij}\) be 1 if \(i = j\) and 0 otherwise, but if we're not fixing a basis then we can't guarantee these values; all we can say is that \(g_{ij} = g_{ji}\).
We also have a similar tensor, living in the dual space, \(g^{ij}\), that is also invariant, where we define \(g^{ij}g_{jk} = I_k^i\).
We also have a similar tensor, living in the dual space, \(g^{ij}\), that is also invariant, where we define \(g^{ij}g_{jk} = I_k^i\).
In fact, given such a pair of tensors, we can actually define \(O(n, \mb k)\) as the largest matrix group in \(End(\mb k^n)\) that leaves these tensors invariant. We don't technically need both tensors; what we do need is for \(g\) to be nondegenerate; in other words, for any \(v \in V\), there has to be some \(u \in V\) such that \(g(u, v)\) is nonzero.
For \(C_n\) we have a similar tensor \(\omega = \omega_{ij}e^i \otimes e^j\), where \(\omega_{ij} = -\omega_{ji}\) and again we demand that \(\omega\) be nondegenerate. We call this the symplectic form. The standard basis has \(\omega(e_i, e_{i + n}) = 1 = -\omega(e_{i+n}, e_i)\) for \(1 \leq i \leq n\), and 0 for all other pairs of indices. We define \(C_n\) to be the largest group that leaves \(\omega\) invariant.
For \(g_{ij}\) and \(\omega_{ij}\), the nondegeneracy allows us to match elements of \(V\) with \(V^\vee\) in a way that is invariant under the respective groups. Normally, for a basis of \(V\) the dual basis transforms in the opposite way so the matching in one basis does not correspond to a matching in another. Hence for most groups, the action on \(V\) and the corresponding action on \(V^\vee\) are inequivalent. But for \(O(n)\) and \(C_n\) they are equivalent.
For \(C_n\) we have a similar tensor \(\omega = \omega_{ij}e^i \otimes e^j\), where \(\omega_{ij} = -\omega_{ji}\) and again we demand that \(\omega\) be nondegenerate. We call this the symplectic form. The standard basis has \(\omega(e_i, e_{i + n}) = 1 = -\omega(e_{i+n}, e_i)\) for \(1 \leq i \leq n\), and 0 for all other pairs of indices. We define \(C_n\) to be the largest group that leaves \(\omega\) invariant.
For \(g_{ij}\) and \(\omega_{ij}\), the nondegeneracy allows us to match elements of \(V\) with \(V^\vee\) in a way that is invariant under the respective groups. Normally, for a basis of \(V\) the dual basis transforms in the opposite way so the matching in one basis does not correspond to a matching in another. Hence for most groups, the action on \(V\) and the corresponding action on \(V^\vee\) are inequivalent. But for \(O(n)\) and \(C_n\) they are equivalent.
Another important tensor is the Levi-Civita tensor, \(\varepsilon\). For a vector space \(V\) of dimension \(n\), \(\varepsilon = \varepsilon_{i_1i_2\ldots i_n}\) lives in \((V^\vee)^{\otimes n}\). It takes in \(n\) vectors and returns the size of the \(n\)-dimensional parallelepiped that those vectors give the edges of. If we work with \(\mb k = \mb R\), we often call the Levi-Civita form the volume form.
Since matrices with determinant 1 don't change the volume of such an object, we get that elements of \(SL(n)\) leave this tensor invariant, and conversely \(SL(n)\) is the set of all matrices that have this as an invariant.
The Levi-Civita tensor is fully antisymmetric, in that swapping any two indices gives you a minus sign.
It turns out that in dimension \(2n\), \(\varepsilon\) can be built out of copies of the \(\omega\) tensor due to the nondegeneracy condition, so \(\varepsilon\) is automatically invariant under \(C_n\).
Since matrices with determinant 1 don't change the volume of such an object, we get that elements of \(SL(n)\) leave this tensor invariant, and conversely \(SL(n)\) is the set of all matrices that have this as an invariant.
The Levi-Civita tensor is fully antisymmetric, in that swapping any two indices gives you a minus sign.
It turns out that in dimension \(2n\), \(\varepsilon\) can be built out of copies of the \(\omega\) tensor due to the nondegeneracy condition, so \(\varepsilon\) is automatically invariant under \(C_n\).
Also recall the map \(Asym_n\) that takes an element of \(V^{\otimes n}\) and returns its antisymmetrization:
$$Asym_2(u\otimes v) = \frac{1}{2}(u \otimes v - v \otimes u)$$
$$Asym_3(u \otimes v \otimes w) = \frac{1}{3!}(u \otimes v \otimes w - u\otimes w\otimes v + v\otimes w\otimes u - \ldots)$$
\(Asym\) can be written in tensor indices, with \(n\) lower indices and \(n\) upper indices. We get that, for \(S_n\) the group of permutations of the numbers \(1,\ldots, n\), we have
\(Asym\) can be written in tensor indices, with \(n\) lower indices and \(n\) upper indices. We get that, for \(S_n\) the group of permutations of the numbers \(1,\ldots, n\), we have
$$Asym_{i_1i_2\ldots i_n}^{j_1j_2\ldots j_n} = \frac{1}{n!}\sum_{\sigma \in S_n} sign(\sigma) I_{i_1}^{j_{\sigma(1)}} I_{i_2}^{j_{\sigma(2)}}\cdots I_{i_n}^{j_{\sigma(n)}}$$
where \(sign(\sigma)\) is the number of swaps it takes to get from the ordering \(1,\ldots n\) to \(\sigma(1),\ldots, \sigma(n)\).
\(\varepsilon\) has a dual such that
\(\varepsilon_{i_1i_2\ldots i_n}\varepsilon^{j_1j_2\ldots j_n} = Asym_{i_1i_2\ldots i_n}^{j_1j_2\ldots j_n}\)
Note that there is no contraction here.
\(Asym_n\) is a tensor invariant for all \(n\), being built from identities. In fact, all of the maps built from partial asymmetrizations and partial symmetrizations are tensor invaraints, and are one way to extract simpler tensor invariants from more complicated ones.
\(Asym_n\) is a tensor invariant for all \(n\), being built from identities. In fact, all of the maps built from partial asymmetrizations and partial symmetrizations are tensor invaraints, and are one way to extract simpler tensor invariants from more complicated ones.
The other simple Lie groups can also be expressed in terms of their invariant tensors for various representations, but these are more complicated.
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