Consider the action of $ GL_d(\mathbb{C}) $ on $ V=\mathbb{C}^d $ and the diagonal action of $ GL_d(\mathbb{C}) $ on $ V^{\otimes n} $ $$ g \cdot \Big(\otimes_{i=1}^n v_i\Big):= g^{\otimes n}\Big(\otimes_{i=1}^n v_i\Big)= \otimes_{i=1}^n g v_i $$ Let $ e_1,\dots , e_d $ be a the standard basis for $ V $. And let the $$ e_{i_1}\otimes \dots \otimes e_{i_n} $$ be the standard tensor basis for $ V^{\otimes n} $. Any vector which is an integral linear combination of the standard tensor basis we will call an integral vector.

Let $ W $ be a $ GL_n(\mathbb{C}) $ sub representation of $ V^{\otimes n} $. By Schur-Weyl duality $ W $ admits a basis of integral vectors.

Let $ G $ be a closed subgroup of $ GL_d(\mathbb{C}) $ which acts irreducibly on $ V $. Consider the diagonal action of $ G $ on $ V^{\otimes n} $. Let $ W $ be a (faithful) $ G $ sub representation of $ V^{\otimes n} $. Does $ W $ always admit an integral basis?

EDIT: Just providing extra details for the awesome example Nate gave. The action of $ A_4 $ on $ V^{\otimes 2} $ decomposes in the following way. The space of all antisymmetric tensors, spanned by $$ e_1 \otimes e_2-e_2 \otimes e_1, e_1 \otimes e_3-e_3 \otimes e_1, e_2 \otimes e_3-e_3 \otimes e_2 $$ is a 3d irrep of $ A_4 $. The space of all off diagonal symmetric tensors, spanned by $$ e_1 \otimes e_2+e_2 \otimes e_1, e_1 \otimes e_3+e_3 \otimes e_1, e_2 \otimes e_3+e_3 \otimes e_2 $$ is also a 3d irrep of $ A_4 $. The span of $$ e_1 \otimes e_1 + e_2 \otimes e_2 + e_3 \otimes e_3 $$ is the trivial irrep of $ A_4 $. The span of $$ e_1 \otimes e_1 + \zeta_3\, e_2 \otimes e_2 + \overline{\zeta_3}\, e_3 \otimes e_3 $$ is the first nontrivial character of $ A_4 $ (I think this is what Nate calls $S^{(2,2)+}$). And the span of $$ e_1 \otimes e_1 + \overline{\zeta_3}\, e_2 \otimes e_2 + \zeta_3\, e_3 \otimes e_3 $$ is the other (complex conjugate) nontrivial character of $ A_4 $ (I think this is what Nate calls $S^{(2,2)-}$).

There is a well known 3d irrep of $ S_4 $ as the group of all signed permutation matrices in $ SO_3 $ and a 3d irrep of $ A_4 $ as the group of signed permutation matrices in $ SO_3 $ with an even number of minus signs. Here $ V^{\otimes 2} $ can be viewed as that copy of $ A_4 $ acting by conjugation on the space of all $ 3 \times 3 $ complex matrices. So the antisymmetric part is $ \mathfrak{so}_3(\mathbb{C}) $. The symmetric off diagonal rep can actually be thought of as symmetric matrices with 0s on the diagonal. The trivial irrep is the scalar matrices and the other two 1d irreps corresponds to the rest of the diagonal matrices.

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    $\begingroup$ I would not call an arbitrary subrepresentation of $V^{\otimes n}$ a "Schur-Weyl type representation," that's much too weak a condition. It should be a representation obtained via an action resembling the group action of $S_n$; e.g. there is a Schur-Weyl duality for quantum groups where $S_n$ is replaced by the braid group $B_n$ and I would consider representations obtained that way "Schur-Weyl type." $\endgroup$ Sep 13, 2022 at 18:01
  • $\begingroup$ Ok I changed the title $\endgroup$ Sep 13, 2022 at 19:57

1 Answer 1


No. Even if $V$ is defined integrally $V^{\otimes n}$ can have complex subrepresentations that can't be defined integrally. In fact, so long as $V$ is faithful (i.e. does not factor through a quotient group) then irreducible every representation of $G$ appears in $V^{\otimes n}$.

As a concrete example: Take the alternating group $A_4$. If we take $V = \mathbb{C}^3$ to be the standard $3$-dimensional irreducible representation. Then $V^{\otimes 2}$ contains two 1-dimensional subrepresentations $S^{(2,2)+}$ and $S^{(2,2)-}$ which are not defined integrally -- they require a third root of unity. Note that for $S_4$ we don't see this, the direct sum of these two form a single irreducible $S_4$ representation $S^{(2,2)}$ that is defined over $\mathbb{Z}$.

  • $\begingroup$ For a reference for the claim at the end of the first paragraph about every irreducible appearing in $V^{\otimes n}$, for finite groups see mathoverflow.net/questions/18194/… and for compact groups the statement needs to be modified but see mathoverflow.net/questions/58633/…. $\endgroup$ Sep 13, 2022 at 19:59
  • $\begingroup$ Looking back I realize that the examples that motivated me to ask this question were all for the case that $ W $ is a faithful representation of $ G $. Do you know any counterexamples for that case that $ W $ is faithful? / Would you mind if I changed my question to add that requirement? $\endgroup$ Sep 13, 2022 at 20:01
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    $\begingroup$ Take $A_5$ instead of $A_4$, with $V = S^{(4,1)}$ the standard 4 dimensional representation and $W = S^{(3,1,1)\pm} \subseteq V^{2}$. $\endgroup$
    – Nate
    Sep 14, 2022 at 12:40

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