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The symmetric group $\mathfrak{S}_n$ can be regarded as a subgroup of the orthogonal group $\textrm{O}(n)$ via the permutation matrices. Let $V$ be a finite dimensional $\textrm{O}(n)$-module and $\varphi: \mathbb{R}^n\to V$ an $\mathfrak{S}_n$-equivariant linear map where $\mathfrak{S}_n$ acts on $\mathbb{R}^n$ in the obvious way. Finally, let $d:\mathbb{R}^n\to\textrm{Sym}_2(\mathbb{R}^n)$ be the map that sends a vector to the corresponding diagonal matrix.

Are there criteria on $\varphi$ that ensure the existence of an $\textrm{O}(n)$-equivariant linear map $\Phi: \textrm{Sym}_2(\mathbb{R}^n)\to V$ such that $\Phi\circ d=\varphi$?

Note that such a map, if it exists, is unique since every real symmetric matrix is diagonalizable with orthogonal matrices.

Edit: This is to show that the map suggested by Aurel is in general not linear. Let $V$ be the representation of $\textrm{O}(n)$ where rotation about angle $t$ acts by $\begin{pmatrix}\cos(6t)&-\sin(6t)\\ \sin(6t)& \cos(6t)\end{pmatrix}$ and reflection at the $x$-axes by $\begin{pmatrix}1&0\\ 0& -1\end{pmatrix}$. Then one can check that the map $\varphi:\mathbb{R}^2\to V$ with $\varphi\binom{1}{1}=0$ and $\varphi\binom{1}{-1}=\binom{1}{0}$ satisfies the conditions described by Aurel. However, $V$ is not part of the decompostion of $\textrm{Sym}_2\mathbb{R}^2$ into irreducibles.

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    $\begingroup$ Do you really seek invariant map or do you actually want equivariant map? $\endgroup$ Jun 24 '20 at 17:02
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    $\begingroup$ Sorry, I mean equivariant. $\endgroup$
    – Hans
    Jun 24 '20 at 17:13
  • $\begingroup$ Here's a proposed answer: The map extends iff the Casimir element in ${U}(\mathfrak{o}_n)$ scales the image of the map $\varphi$ by the right scalar. $\endgroup$
    – Nate
    Jun 25 '20 at 17:14
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If I have understood the problem correctly, the map $\Phi$ deterimines $\varphi = \Phi \circ d$, so the question amounts to classifying possible compositions $\Phi \circ d$, where $d$ is the "diagonal" map, and $\Phi$ is $O(n)$ equivariant. Clearly the image of $\varphi$ must be contained in the image of $\Phi$ which is isomorphic to a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$.

Let's begin by noting that as a representation of $\mathfrak{S}_n$, we have the decomposition $\mathbb{R}^n = \mathbf{1}_{\mathfrak{S}_n} \oplus U$, where $\mathbf{1}_{\mathfrak{S}_n}$ is the trivial representation (spanned by the "all ones" vector), and $U$ is the (irreducible) standard representation (consisting of "mean zero" vectors).

Similarly, as a representation of $O(n)$, the representation $\mathrm{Sym}_2(\mathbb{R}^n)$ (which we are interpreting as symmetric $n \times n$ matrices under conjugation) decomposes as $\mathbf{1}_{O(n)} \oplus W$, where $\mathbf{1}_{O(n)}$ is the trivial representation (spanned by the identity matrix), and $W$ is an irreducible representation which consists of symmetric matrices of trace zero.

It is not difficult to see that the map $d: \mathbf{1}_{\mathfrak{S}_n} \oplus U \to \mathbf{1}_{O(n)} \oplus W$ is of the form $f \oplus g$, with $f : \mathbf{1}_{\mathfrak{S}_n} \to \mathbf{1}_{O(n)}$ (it sends the all-ones vector to the identity matrix) and $g: U \to W$.

Now, $\Phi$ must map $\mathbf{1}_{O(n)}$ to an invariant vector (possibly zero), and it must map $W$ to either a copy of $W$ or zero. So, we must give an "intrinsic" description of the image of $g$, i.e. traceless diagonal matrices, inside $W$.

Note that $O(n)$ contains not only $\mathfrak{S}_n$, but the larger hyperoctahedral group $\mathfrak{H}_n = C_2 \wr \mathfrak{S}_n = C_2^n \rtimes \mathfrak{S}_n$, via signed permutation matrices (permutation matrices, but the nonzero entries can be $\pm 1$). The hyperoctahedral group contains a subgroup $C_2^n$ consisting of diagonal matrices with entries $\pm 1$. It is not difficult to check that a matrix that commutes with every element of $C_2^n$ (viewed as a diagonal matrix) must itself be diagonal, and conversely it is clear that every diagonal matrix commutes with $C_2^n$. Phrasing this in terms of the module structure (commuting with = fixed under conjugation by), the diagonal matrices are fixed by the action of $C_2^n$.

We are now in a position to give the characterisation. A map $\varphi: \mathbb{R}^n \to V$ may be written in the form $\Phi \circ d$ if and only if all of the following conditions are satisfied.

    1. The $O(n)$-module generated by the image of $\varphi$ is a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$.
    2. The image of the "all ones" vector is $O(n)$ invariant (possibly zero).
    3. The image of $\varphi$ is fixed pointwise by $C_2^n$ (and therefore has a $\mathfrak{H}_n$ action that factors through $\mathfrak{S}_n$).

We've demonstrated that these conditions are necessary, so let's check that they are also sufficient. Note that by (2), given $\varphi$ as above, the restriction $\mathbf{1}_{\mathfrak{S}_n} \to V$ determines the restriction $\Phi: \mathbf{1}_{O(n)} \to V$. So it suffices to construct the restriction $\Phi: W \to V$. By (1), we are guaranteed to find such a map whose image agrees with the $O(n)$-module generated by $\varphi(U)$; we check that that (3) implies that $\varphi(U)$ coincides with the image of the (traceless) diagonal matrices in $W$ (because these form an irreducible representation of $\mathfrak{S}_n$, rescaling $\Phi$ is all that is needed to guarantee pointwise agreement of $\varphi$ and $\Phi \circ d$).

As a representation of $\mathfrak{H}_n$, $\mathrm{Sym}_2(\mathbb{R}^n)$ can be decomposed as follows. Recognise that

$$ \mathbb{R}^n = \mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}), $$

where $\varepsilon$ is the sign character of $\mathfrak{H}_1 = C_2$. Now, using a little Mackey theory (to write down the tensor square, and then extract they symmetric part), we see that

$$ \mathrm{Sym}_2(\mathbb{R}^n) = \mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathrm{Sym}_2(\varepsilon) \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}) \oplus \mathrm{Ind}_{\mathfrak{H}_2 \otimes \mathfrak{H}_{n-2}}^{\mathfrak{H}_n}(\varepsilon \otimes \varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}) $$

where $\varepsilon \otimes \varepsilon = \varepsilon \otimes \varepsilon \otimes \mathbf{1}_{\mathfrak{S}_2}$ is a one-dimensional representation of $\mathfrak{H}_2$ where a signed matrix acts by $(-1)^m$, where $m$ is the number of $-1$'s in the signed matrix. Enthusiasts of wreath products will recognise both summands as being irreducible representations of the hyperoctahedral group. The key point here is that $\mathrm{Sym}_2(\varepsilon) = \mathbf{1}_{C_2}$, so the first summand simply becomes

$$ \mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathbf{1}_{\mathfrak{H}_1} \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}), $$

which is simply $\mathbb{R}^n$, viewed as a $\mathfrak{H}_n$ module that factors through the action of $\mathfrak{S}_n$. In particular, this summand corresponds to diagonal matrices in $\mathrm{Sym}_2(\mathbb{R}^n)$. (The other summand does not factor through the $\mathfrak{S}_n$ action.)

One final comment: If $n \geq 4$, as a representation of $\mathfrak{S}_n$,

$$ \mathrm{Sym}_2(\mathbb{R}^n) = S^{(n)} \oplus S^{(n)} \oplus S^{(n-1,1)} \oplus S^{(n-1,1)} \oplus S^{(n-2,2)} $$

where $S^\lambda$ is a Specht module (irreducible representation) indexed by $\lambda$. The key point here is that $U = S^{(n-1,1)}$ appears with multiplicity 2. This means that trace zero symmetric matrices are not the unique subspace of $\mathrm{Sym}_2(\mathbb{R}^n)$ isomorphic to $U$. This means that condition (3) is not automatic. In particular, Nate's suggestion of understanding the interaction with the Casimir element will help pin down condition (1) (it helps identify the ambient $O(n)$ representation), but some extra information will be required to detect condition (3).

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  • $\begingroup$ Thanks for that wonderful answer! A last question regarding the Casimir element: I understand that the irreducible components of a representation will be eigenspaces of the Casimir operator. But is it really clear that non-isomorphic irreducibles will correspond to different eigenvalues? $\endgroup$
    – Hans
    Jun 27 '20 at 9:34
  • $\begingroup$ @Christopher The second point is always satisfied, no? The $\mathfrak{S}_n$-homomorphism $\varphi$ sends invariant vectors to invariant vectors. I have trouble following your argument. How do you use $\mathfrak{h}_n$-module description? $\endgroup$ Jun 27 '20 at 11:41
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    $\begingroup$ @Hans It's not clear and maybe it's even not true. Casimir operator acts on a highest weight module by a scalar that can be calculated from the highest weight. You'd need to check whether this scalar uniquely determines the representation $\mathrm{Sym}_2^0.$ $\endgroup$ Jun 27 '20 at 11:42
  • $\begingroup$ Thanks @VítTuček. Regarding the second point: For example the 2x2 matrix $\binom{01}{10}$ is $\mathfrak{S}_2$ invariant but not $\textrm{O}(2)$ invariant. So (2) is not automatically true. $\endgroup$
    – Hans
    Jun 27 '20 at 13:29
  • $\begingroup$ You have already answered each other's questions, but I just wanted to mention that condition (2) is necessary because the image of $\mathbf{1}_{\mathfrak{S}_n}$ is a priori a trivial representation of $\mathfrak{S}_n$ only. But there are many irreducible representations of $O(n)$ (including $W$) that contain $\mathbf{1}_{\mathfrak{S}_n}$ when restricted to $\mathfrak{S}_n$. It is because $d$ maps $\mathbf{1}_{\mathfrak{S}_n}$ to $\mathbf{1}_{O(n)}$ (and not a $\mathfrak{S}_n$ trivial contained in $W$) that condition (2) arises. $\endgroup$ Jun 27 '20 at 16:32
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Not an answer, just too long for a comment.

All representations in sight can be decomposed as direct sums of irreducible representations. By the Schur lemma, mapping between irreducibles can be either zero, or multiple of identity. In this way the problem can be broken down into subproblems. (I am not sure it helps much, but at least you can see immediately, once you have the decomposition, whether there is any $O(n)$ map at all.)

The symmetric matrices decompose as $O(n)$-modules into traceless matrices and multiples of identity matrix. I.e. $$A \mapsto (A - \frac{1}{n}(\mathrm{Tr}\, A)\; \mathrm{Id}) \oplus \frac{1}{n}(\mathrm{Tr}\, A ) \; \mathrm{Id}.$$

If $V$ is a tensor representation (i.e. subrepresentation of $\bigotimes^k\mathbb{R}^n$) then the decomposition into $O(n)$-modules works in a similar way. One substracts all possible traces over all possible pairs of indices and then decomposes resulting modules according to their $\mathfrak{S}_n$-symmetries in indices. (See Goodmann, Wallach for reference.)

Unfortunately, I am not much familiar with representations of finite groups so I don't know what the decomposition of the symmetric matrices looks like.

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Edit: this answer overlooks the requirement that $\Phi$ be linear.

For $a$ a diagonal matrix, let $Z_{O(n)}(a)$ be its centraliser in $O(n)$, i.e. $\{k\in O(n) \mid kak^{-1} = a\}$. This is easy to describe: partition $\{1,\dots,n\}$ into subsets of size $n_1,\dots,n_m$ according to where the components of $a$ are equal; then $Z_{O(n)}(a)$ is a corresponding copy of $O(n_1)\times\dots\times O(n_m)$ (blockwise diagonal if the components of $a$ are sorted).

A necessary condition for the existence of $\Phi$ is that for all $a$, we must have $\varphi(a) \in V^{Z_{O(n)}(a)}$. Simply write down what the equivariance says.

Let's prove that this condition is also sufficient, so assume it is satisfied. Let $s$ be a symmetric matrix, and write it as $kak^{-1}$ where the entries of $a$ are nondecreasing and $k\in O(n)$. Define $\Phi(s) = k\cdot \varphi(a)$. This is well-defined: $a$ depends only on $s$ and the only possible decompositions of $s$ with this $a$ are the $kza(kz)^{-1}$ with $z\in Z_{O(n)}(a)$. The map $\Phi$ is $O(n)$-equivariant by construction (we defined it on each orbit by taking a representative and applying the action). Finally, $\Phi$ agrees with $\varphi$ on all diagonal matrices because of the assumption of permutation equivariance.

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  • $\begingroup$ Thanks for the answer. Could you also explain why $\Phi$ is a linear map? $\endgroup$
    – Hans
    Jun 25 '20 at 7:05
  • $\begingroup$ Oh, I missed that requirement! I'll have to check whether it is. $\endgroup$
    – Aurel
    Jun 25 '20 at 8:47
  • $\begingroup$ I have added a counterexample where $\Phi$ is not linear. $\endgroup$
    – Hans
    Jun 25 '20 at 12:31
  • $\begingroup$ Yeah, that was likely. Thanks for the counterexample. I'll think about the correct problem. $\endgroup$
    – Aurel
    Jun 25 '20 at 14:53

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