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I am interested in applying Schur-Weyl duality to quantum information theory, specifically "qubits". But I have been stuck for some time on understanding how the Young symmetrizers work in this situation.

Let $V$ be a 2 dimensional complex vector space, a qubit. Let $S_3$ be the permutation group on three things, acting on $V\otimes V\otimes V$ by permuting the factors. Write a basis for $V$ as $v_0$ and $v_1$, and the corresponding basis of $V\otimes V\otimes V$ as $v_{000}, v_{001},...,v_{111}$. It's not too hard to see that this representation of $S_3$ is 8 dimensional, and breaks into 4 copies of the trivial representation:

\begin{align*} &v_{000},\\ &v_{100} + v_{010} + v_{001},\\ &v_{011} + v_{101} + v_{110},\\ &v_{111}, \end{align*} and two copies of the 2 dimensional representation: \begin{align*} &\{v_{100}-v_{010}, 2v_{001}-v_{010}-v_{100}\},\\ &\{v_{011}-v_{101}, 2v_{110}-v_{101}-v_{011}\}. \end{align*} Here I have written a basis for each of the 2 dimensional representations. So far so good. But now things start to get murky for me. I would like to produce these 2 dimensional representations via the action of the Young symmetrizer. This operator, $P$ is given by $$ P v_{abc} = v_{abc} + v_{bac} - v_{cba} - v_{cab}. $$ But feeding any of $v_{100}, v_{010}, v_{001}$ into $P$ results in a multiple of $v_{100}-v_{001}.$ For example $$ P v_{100} = v_{100} + v_{010} - v_{001} - v_{010} = v_{100} - v_{001}. $$ I have tried many variations on the definition of $P$, and also letting it act on "the left or the right" of the $v_{abc}.$ Nothing I do seems to produce more than a one dimensional subspace of the sought-after two dimensional space.

What I can do is produce the Specht modules from the group algebra of $S_3$ using the Young symmetrizers, but somewhere between the group algebra and $V\otimes V\otimes V$ I appear to be getting lost.

glS
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Simon Burton
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    The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for. – Joppy Nov 21 '18 at 12:57
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    What you need to do is add in the Schur-Weyl duality part, and use the action of the $\operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity. – Joppy Nov 21 '18 at 12:58
  • Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well. – Simon Burton Nov 22 '18 at 14:36
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    So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $\mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes. – Joppy Nov 22 '18 at 15:00
  • Ah right, I think I am catching up now! The Young symmetrizer above gives me the two dimensional space spanned by $v_{100}-v_{001}$ and $v_{011}-v_{110}$ which is indeed an irreducible sub-representation of the $GL(n, C)$ action on $V\otimes V\otimes V.$ Thankyou for your patience. – Simon Burton Nov 22 '18 at 15:10
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    In fact, the image of any element of $\mathbb{C}[S_n]$ acting on $V^{\otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type. – Joppy Nov 22 '18 at 15:27
  • @Joppy Did you mean to say "isomorphic to the two-dimensional irrep of $S_3$"? Because I think what Simon is pointing out is that you're not getting an $S_3$ module (irreducible or otherwise). For, if the image under this Young symmetrizer were closed under the action of $S_3$, then the intersection with the $S_3$-submodule spanned by $v_{001}, v_{010}, v_{100}$ would also be closed under the $S_3$-action. Which it manifestly isn't because that intersection is just the 1-dimensional space spanned by $v_{100} - v_{001}$. – user43208 Nov 23 '18 at 15:25
  • @user43208 Ah yes, I was getting confused. Since the young projectors don’t commute with much of the symmetric algebra, after using them you get a submodule for a different action. If using them on $\mathbb{C}[S_n]$, you can project using the right action, and still have the left action remaining. But for $V^{\otimes n}$, after projecting there is only the $GL(V)$ action left. – Joppy Nov 23 '18 at 23:38
  • @Joppy Thanks, Joppy. Maybe you know the answer to the following question. Suppose we take the sum of Young symmetrizers, indexed over all possible tableaux of a given Young diagram $D$ of size $n$. Do we get a scalar multiple of the primitive central idempotent in the group algebra $\mathbb{C}[S_n]$, the one that corresponds to the identity on the irrep $V_D$ classified by $D$ when we consider the algebra isomorphism $\mathbb{C}[S_n] \cong \prod_D \hom(V_D, V_D)$? I found this was the case for $S_3$, and conjecture it's true for $S_n$ generally. – user43208 Nov 24 '18 at 01:22
  • The reason I ask is that this might provide an acceptable answer for the OP: that it's the sum of the symmetrizers for given $D$ that gives the operator that projects a given rep onto its $V_D$ component (counted with multiplicity, of course). – user43208 Nov 24 '18 at 01:27
  • @user43208 the answer is yes: $\sum_{\pi \in S_k} \pi c_\lambda \pi^{-1}$ yields a primitive central idempotent, that is, the projector onto the isotypic component. Also see the question https://math.stackexchange.com/questions/3345754/constructing-a-centrally-primitive-idempotent-in-the-group-algebra-of-the-symmet – Felix Huber Nov 15 '19 at 19:33

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I'll try to expand upon my answers in the comments here. You have the $\mathbb{C}$-vector space $V = \{v_0, v_1\}$, and the tensor product $V^{\otimes 3}$ has both a left action by $\operatorname{GL}(V) \cong \operatorname{GL}_2(\mathbb{C})$, $g \cdot (v \otimes u \otimes w) = gv \otimes gu \otimes gw$, and a right action by $S_3$, given by the formula $(u_1 \otimes u_2 \otimes u_3) \cdot \sigma = u_{\sigma(1)} \otimes u_{\sigma(2)} \otimes u_{\sigma(3)}$.

Now, by only viewing $V^{\otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations: $$\begin{aligned} &v_{000}, \\ &v_{100} + v_{010} + v_{001}, \\ &v_{110} + v_{101} + v_{011}, \\ &v_{111} \end{aligned}$$ and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$: $$\begin{aligned} &v_{100} - v_{010}, \\ &v_{001} - v_{010}, \\ &v_{011} - v_{101}, \\ &v_{110} - v_{101} \end{aligned}$$

I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $\mathbb{C}[S_3]$ (they commute with everything).

One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $\operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $\operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by $\{v_{100} - v_{010}, v_{001} - v_{010}\}$ is highest-weight of weight $(2, 1)$, and so this $\operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.

It is possible to further decompose these spaces by making more choices in either $\operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $\operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.

Joppy
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    A small comment to "you could also make choices of Young symmetrisers I guess?": Different Young Symmetrizers yield different irreps. But only up to $n<=4$ will the Young Symmetrizers give a complete set of mutually orthogonal primitive idempotents. For higher $n$ this approach fails, and thus the isotypic component doesn't decompose this nicely. There are methods to fix this however, see e.g. https://arxiv.org/abs/1610.10088. – Felix Huber Nov 15 '19 at 19:49