# Which representations of $\mathfrak{sl}(2)$ are homomorphic images of the tensor product of finitely many copies of $\mathbb{C}^2$?

My questions may turn out to be related to Schur functors.

If $$\mathfrak{g}$$ is a complex semisimple Lie algebra and $$\lambda$$ is the highest weight of an irreducible representation $$V$$ of $$\mathfrak{g}$$, I am interested in restricting the representation to an $$\mathfrak{sl}(2)$$ subalgebra of $$\mathfrak{g}$$. For example, if $$\mathfrak{g} = \mathfrak{sl}(n)$$, then I am interested in $$V = \mathbb{C}^n$$ being the standard representation. If $$\mathfrak{g} = \mathfrak{so(n)}$$, I am interested in $$V = \mathbb{C}^n$$, i.e. in the vector representation. Similarly if $$\mathfrak{g} = Sp(m, \mathbb{C})$$, I am interested in $$V = \mathbb{C}^{2m}$$ being its standard representation. I am also interested in the $$5$$ exceptional cases and their fundamental representations.

As for the $$\mathfrak{sl}(2)$$ subalgebra of $$\mathfrak{g}$$, I am mostly interested in "the" so-called "regular" $$\mathfrak{sl}(2)$$ subalgebra (it is unique up to conjugation in $$\mathfrak{g}$$), which was studied for instance by Kostant and others.

I can now formulate my questions. From now on, consider the representation $$V$$, but view it as a representation of a fixed $$\mathfrak{sl}(2)$$ subalgebra of $$\mathfrak{g}$$ (for the sake of this post, one may for instance assume it to be a regular $$\mathfrak{sl}(2)$$ subalgebra).

Question 1: Does there always exist a positive integer $$m$$, and a $$\mathfrak{sl}(2)$$ intertwining (meaning the corresponding notion at the Lie algebra level) linear map

$$\mathbb{C}^2 \otimes \cdots \otimes \mathbb{C}^2 \to V,$$

which is onto, where the domain is the tensor product of $$m$$ copies of $$\mathbb{C}^2$$?

Question 2: Assuming the answer to question 1 is yes, can one write down such a linear map using weights and roots? For example, what would $$m$$ be in terms of Lie-theoretic data, etc?

Note: I did edit my post quite a bit, so some of the comments may not make sense now to a new reader, because of my edits. The new reader should keep this in mind. However, these are now the precise questions that I would like to see answered. I apologize for the little mess that I have made, with my latest edits. I think though that my questions are more focused now, in this version.

• Does your "regular" $\frak{sl}_2$ subalgebra correspond to the half-sum of positive roots $\rho$ in the weight lattice of $\frak{g}$ ? May 4 at 21:19
• I do not recommend using the term "symmetric/anti-symmetric tensor product", which is confusing and non-standard, since there is no such thing as a symmetric or anti-symmetric tensor product of two different representations. Write "symmetric or exterior power" instead. May 4 at 21:35
• @DmitryVaintrob, I edited the post a little, taking into account your comment. You are right of course. May 4 at 21:46
• If I remember correctly, the spin-representation of $\frak{so}(2n)$ as a vector spaces is just the exterior algebra $\bigwedge^\bullet \Bbb{C}^n$ and, upon restriction to this "regular" $\frak{sl}_2$ subalgebra, $H = \exp(h)$ acts as Berezin-Fourier transform $\eta \mapsto \int d \theta_1 \wedge \cdots d\theta_n \, \eta \exp \big( \underline{\theta}^T J \underline{\sigma} \big)$ from which you can write down the character and check whether or not it factorizes into a product of (smaller) characters. I believe it does ---I did this computation ten years ago but I've lost my notes :/ May 4 at 22:06
• Look at lecture 20 in Fulton & Harris. I'm sure what I wrote above has errors. If you choose a basis $\theta_1, \dots, \theta_n$ of $\Bbb{C}^n$ then the Berezin-Fourier transform is just $\theta_I \mapsto \theta_{\overline{I}}$ where $I = \{i_1 < \cdots < i_k \}$ is a $k$-element subset of $\{1, \dots, n \}$ and $\overline{I}$ is its complement and $\theta_I = \theta_{i_i} \wedge \cdots \wedge \theta_{i_k}$. May 4 at 22:16

$$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sym{Sym}$$Question: "Which representations of $$\mathfrak{sl}(2)$$ are homomorphic images of the tensor product of finitely many copies of $$\mathbb{C}^2$$?"

Answer: If $$k$$ is the field of complex numbers and if $$W:=k\{e_1,e_2\}$$ it follows that any finite dimensional irreducible $$\SL(W)$$-module $$V$$ decomposes as $$V \cong \bigoplus_i \Sym_k^{m_i}(W),$$ where $$m_i\geq 1$$ are integers. There is for every $$i$$ a surjection $$\pi_i: W^{\otimes m_i} \rightarrow \Sym_k^{m_i}(W)$$ hence you get a surjection $$\pi: U:=\bigoplus_i W^{\otimes m_i} \rightarrow V \cong \bigoplus_i \Sym_k^{m_i}(W).$$ Hence every module $$V$$ is a quotient of a direct sum of tensor powers of $$W$$. I'm unsure if you can choose $$U$$ to be a tensor product of $$W$$ — you may need to take direct sums.

In Fulton–Harris's book "Representation theory - a first course", §15.3, they give an elementary construction of all finite dimensional irreducible $$\SL(n,k)$$-modules as submodules of the tensor product $$\Sym^{a_1}(V)\otimes \Sym^{a_2}({\bigwedge}^2 V) \otimes \dotsb \otimes \Sym^{a_{n-1}}({\bigwedge}^{n-1} V)$$ where $$V:=k^n$$, by giving an explicit construction of the highest weight vector. In the general case you must use tensor and exterior products to get all representations.

Note 1: For $$\SL(2,k)$$ it follows $$W \cong W^*$$ is an isomorphism, hence you get isomorphisms of $$\SL(2,k)$$-modules $$W^{\otimes m} \cong \operatorname{End}_k(W)^{\otimes n}$$ if $$m=2n$$ and $$W^{\otimes m} \cong \operatorname{End}_k(W)^{\otimes n}\otimes W$$ if $$m=2n+1$$.

Note 2: On page 472 in FH you find the following construction in characteristic zero: There is for any $$n,d\geq 1$$ a split surjection $$V^{\otimes dn} \xrightarrow p \Sym^n({\bigwedge}^d V)$$ and if $$G:=\SL(V)$$ it follows $$p$$ is a split surjective map of $$G$$-modules. Using this construction you get for any integers $$a_1,\dotsc,a_{n-1}$$ a split surjection of $$G$$-modules $$V^{\otimes \sum_i ia_i} \rightarrow \bigotimes_i \Sym^{a_i}({\bigwedge}^i V)$$ and since any finite dimensional irreducible $$G$$-module $$W$$ is a submodule of $$\bigotimes_i \Sym^{a_i}({\bigwedge}^i V)$$ it follows you get a split surjection of $$G$$-modules $$V^{\otimes \sum_i ia_i} \rightarrow W$$ for any finite dimensional irreducible $$G$$-module $$W$$. Hence in characteristic zero any finite dimensional irreducible $$W$$ may be realized as a submodule (or quotient module) of $$V^{\otimes n}$$ for some integer $$n$$. Hence if you want to give an explicit and elementary construction of all finite dimensional irreducible $$\SL(V)$$-modules, you may use the tensor powers $$V^{\otimes n}$$ for all $$n\geq 1$$. The above argument gives a construction of a highest weight vector

$$v_{\lambda} \in V^{\otimes n}$$

for any finite dimensional irreducible $$SL(V)$$-module $$V(\lambda)$$.

Example: Given $$W_1\subseteq V^{\otimes n_1}, W_2 \subseteq V^{\otimes n_2}$$. Let $$V:=\mathbb{C}^{n}$$ and let $$T:=\wedge^n V$$ be the trivial module. You may construct $$T \subseteq V^{\otimes n}$$ as a sub-module. For $$W_1\oplus W_2$$ to be realized as a quotient/submodule of $$V^{\otimes d}$$ for some $$d$$ there are obvious "obstructions". If $$n_1,n_2$$ are even and $$n$$ is odd there is no common tensorpower $$V^{\otimes d}$$ containing the module

$$W_1 \otimes T^{\otimes i}\text{ and }W_2\otimes T^{\otimes j}$$

for integers $$i,j$$. This is similar to the case of $$SL(\mathbb{C}^2)$$.

"Thank you so much for the reference to that fact in Fulton/Harris. I will check it out. – Malkoun 18 hours"

The above construction gives an "elementary" construction of all finite dimensional irreducible $$\SL(V)$$-modules by constructing an explicit highest weight vector $$v\in V^{\otimes n}$$ for some $$n \geq 1$$. I believe one of the reasons for introducing the Schur-Weyl functors in FH is to relate this study to the study of the representations of the symmetric group and combinatorics.

"I am still reading parts of Fulton and Harris. I get that such "tensorial" constructions can produce irreducible representations for the An type Lie groups, via the works of Schur, Weyl, Young etc. I wonder if similar constructions exist for all classical groups (I suspect the answer is yes). But what about the 5 exceptional cases? Is there a known general explicit construction for the highest weight representations which includes such "tensorial" constructions, but also works for the exceptional Lie groups? If this is known, then this would probably be directly relevant to my 2 projects! – Malkoun"

Dixmiers book "Enveloping algebras" give a construction of all finite dimensional irreducible $$\mathfrak{g}$$-modules for any semi simple Lie algebra using the universal enveloping algebra $$U(\mathfrak{g})$$ and Verma modules. Any finite dimensional irreducible $$\mathfrak{g}$$-module $$V(\lambda)$$ has a "canonical" highest weight vector $$v$$ and you get an exact sequence

$$0 \rightarrow I(\lambda) \rightarrow U(\mathfrak{g}) \rightarrow^{\rho_v} V(\lambda) \rightarrow 0$$

where $$v\in V(\lambda)$$ is a highest weight vector for $$V(\lambda)$$ and the map $$\rho_v$$ is defined by $$\rho_v(x):=xv$$. The ideal $$I(\lambda)$$ is the (2-sided) "annihilator ideal" of the vector $$v$$. In Theorem 7.2.6 in Dixmier they describe all finite dimensional irreducible $$\mathfrak{g}$$-modules by giving an explicit 2-sided maximal ideal $$I(\lambda) \subseteq U(\mathfrak{g})$$. In Proposition 7.2.7 they give generators of $$I(\lambda)$$.

• yes, you are right. But it is important for my purposes not to use direct sums. I may edit more a bit later to explain what I have in mind. May 12 at 12:40
• Thank you so much for the reference to that fact in Fulton/Harris. I will check it out. May 12 at 13:39
• @hm2020 I have rolled back the complete deletion of this answer. It is not accepted practice here to delete one's posts after deciding to leave the community. The preferred way is described in the help center. If you do that, all your posts will stay, but will be made anonymous, no longer connecting them to your previous account. You are of course free to edit out personal information before that, if you think there is any, but the maths should stay. Aug 6 at 8:10
• @hm2020 Your reasons for leaving are not relevant to the question whether or not it is appropriate to remove your answers from the site. It is expected of you (and you agreed to that back when you joined MO!) that you leave the answers up for future readers of the question. I recommend to ask on meta if you wish to discuss this policy. Aug 6 at 9:05
• Neither is a sign of disrespect, because neither is about you, only the content of your posts. And again, I urge you to take this to meta. The comments under this answer are supposed to be about this specific answer, not the MO community's attitude towards editing in general. Aug 6 at 9:20

$$\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\sl{\mathfrak{sl}}$$In the comments to hm2020's answer, the OP explains that they want a surjection from $$(\mathbb{C}^2)^{\otimes m}$$, not from a direct sum $$\bigoplus_i (\mathbb{C}^2)^{\otimes m_i}$$. In that case, there is one obstruction: The weights of the representation must be either all even or all odd, and then the answer is yes.

Notation: Let $$h = \left[ \begin{smallmatrix} 1&0 \\0&-1 \end{smallmatrix} \right]$$. Recall that the eigenvalues of $$h$$ on any (finite dimensional) $$\sl(2)$$ representation are integers. We write $$V_i$$ for the $$i$$-eigenspace of $$h$$ on a representation $$V$$, and write $$\chi_V = \sum_i \dim V_i q^i$$ for a formal variable $$q$$. We call $$\chi_V$$ the character of $$V$$.

We put $$R = \mathbb{C}^2$$ with the obvious $$\sl(2)$$ action, and put $$S_k:=\operatorname{Sym}^k R$$ (so $$R = S_1$$).

It is well known that the finite dimensional representation theory of $$\sl(2)$$ is semisimple, and that the simple representations are the $$S_k$$ with character $$q^k + q^{k-2} + \dotsb + q^{-k}$$. Meanwhile, the character of $$R^{\otimes m}$$ is $$(q+q^{-1})^m$$. We thus see that $$\chi_{R^{\otimes m}}$$ is either an even or odd Laurent polynomial in $$q$$, according to the parity of $$m$$, and that $$\chi_{S_k}$$ is either even or odd according to the parity of $$k$$. So $$S_k$$ can only be a summand of $$R^{\otimes m}$$ if $$m \equiv k \bmod 2$$.

Thus, a representation $$\bigoplus S_{k_i}$$ is only a summand of $$R^{\otimes m}$$ if all the $$k_i$$ are congruent modulo $$2$$.

This is also a sufficient criterion. The multiplicity of $$S_k$$ in $$R^{\otimes (k+2 \ell)}$$ is $$\binom{k+2\ell}{\ell} - \binom{k+2\ell}{\ell-1}$$, which goes to infinity as $$\ell$$ goes to infinity. So, for any fixed list of $$k_i$$ which are all congruent to a fixed value $$k$$ modulo $$2$$, the multiplicities of $$S_{k_i}$$ in $$R^{\otimes k+2 \ell}$$ will get arbitrarily large. Thus, any finite dimensional representation obeying this partiy condition will eventually be a summand of $$R^{\otimes m}$$.

I find this more intuitive with the Lie group $$SL(2)$$ instead of the Lie algebra: $$\chi_V(q)$$ is the trace of the Lie group element $$\left[ \begin{smallmatrix} q & 0 \\ 0 & q^{-1} \\ \end{smallmatrix} \right]$$, and the parity condition is that $$\left[ \begin{smallmatrix} -1 & 0 \\ 0 & -1 \\ \end{smallmatrix} \right]$$ must act either $$1$$ or by $$-1$$.

• Thank you so much. Very nice answer. How did you calculate the multiplicity of $S_k$ in $R^{\otimes(k+2l)}$ by the way? Is it by taking the inner product of the 2 characters? May 12 at 13:36
• Essentially, yes. More carefully, I used the Weyl character formula: $\chi_{S_k} = \tfrac{q^{k+1}-q^{-(k+1)}}{q-q^{-1}}$. So the multiplicity of $S_k$ in $V$ is the coefficient of $q^{k+1}$ in $(q-q^{-1}) \chi_V$. Multiply out $(q-q^{-1}) (q+q^{-1})^{m+2 \ell}$ by the binomial theorem to get the claim. May 12 at 13:39
• This answers my question. But let me ask you one more question please. Is there a nice class of irreducible representations $V$ of $\mathfrak{g}$, such that when you restrict $V$ to a regular $sl(2)$-subalgebra of $\mathfrak{g}$, $V$ decomposes into a direct sum of $S_k$s, with all $k$'s having the same parity? Is such a property perhaps true for all irreducible representations $V$? Or at least is it true for minuscule or quasi-minuscule representations? May 12 at 14:12
• Yes, this is true of all (finite dimensional) irreps. All the irreps are the $S_k$'s listed. The only minuscule representation of $\sl(2)$ is $S_1$, and the only quasi-minuscule (but not minuscule) one is $S_2$. You seem like you are not very used to think about $\sl(2)$ representations; perhaps read the early chapters of Fulton and Harris or a similar book? May 12 at 15:21
• I meant that $\mathfrak{g}$ is a complex semisimple Lie algebra (not necessarily $\mathfrak{sl}(2)$) and $V$ is an irreducible representation of $\mathfrak{g}$, which you then view, by restriction, as a representation of a fixed regular $\mathfrak{sl}(2)$ subalgebra of $\mathfrak{g}$. $V$ then decomposes as a direct sum of $S_k$s. My question is whether or not the $k$s necessarily have the same parity. Or if there is a nice class or irreducible representations $V$ of $\mathfrak{g}$ for which the corresponding $k$s necessarily have the same parity. May 12 at 15:32