The context for this question is that I am trying to determine the Grothendieck group of finite-dimensional complex representations of $T = (\mathbb{C}^*)^n$, where $\mathbb{C}^*$ denotes the multiplicative group of complex numbers.

I would like a push toward the answer to this question: let $G$ be a group. Let $\rho:G\to GL(V)$, $\pi: G \to GL(W)$, and $\sigma: G\to GL(U)$ be three finite-dimensional representations, and consider the (not necessarily commutative) diagram of vector spaces and vector space homomorphisms

$$ \require{AMScd} \begin{CD} 0 @>>> U @>{\alpha}>> V @>{\beta}>> W @>>> 0\\ \ @V{\sigma_g}VV @V{\rho_g}VV @V{\pi_g}VV \\ 0 @>>> U @>{\alpha}>> V @>{\beta}>> W @>>> 0 \end{CD} $$ What are necessary and/or sufficient conditions on $\alpha$ and $\beta$ such that the above diagram commutes for all $g\in G$ and the rows are exact? If there is no result leading toward an answer for general groups $G$, then what about when $G=T$?

Relevant: the finite-dimensional representations of $T$ factor into representations of the form $\mathbb{C}_{A_1}\oplus\cdots\oplus\mathbb{C}_{A_m}$, $A_i = (a_1,\ldots,a_n)\in\mathbb{Z}^n$, where $\mathbb{C}_{A_i}$ is the one-dimensional representation given by the linear action $$T\times \mathbb{C}\to\mathbb{C}: t.z = t_1^{a_1}t_2^{a_2}\cdots t_n^{a_n} z.$$

I saw that if either row is exact then the other is automatically exact, and that both rows must split since we're working in finite-dimensional vector spaces, so $\dim U + \dim W = \dim V$. Otherwise I'm not sure how to proceed - I'm a beginner when it comes to this sort of algebra.

  • $\begingroup$ Focus on T (you are not going to be able to say anything in the general case) Since you know all the finite dimensional representations, you can simply look for all short exact sequences! $\endgroup$ – Mariano Suárez-Álvarez Jun 8 '14 at 3:45
  • $\begingroup$ But how do I know which short exact sequences make the diagram commute? Or perhaps a better, smaller question, how do I know which $\alpha$ make the left square commute? $\endgroup$ – Gyu Eun Lee Jun 8 '14 at 3:48
  • $\begingroup$ Consider two representations, each with its decomoosition as a direct sum of 1-dimensional subrepresentations. Find all morphisms of representations betweem them. $\endgroup$ – Mariano Suárez-Álvarez Jun 8 '14 at 3:50
  • $\begingroup$ This is what I see so far: $\sigma_t$ and $\rho_t$ can be thought of as diagonal matrices for each $t\in T$. We are looking for commutativity of matrices $\rho_t\alpha = \alpha\sigma_t$. $\rho_t$ multiplies each row of $\alpha$ by some $\prod t_i^{a_i}$, and $\sigma_t$ multiplies each column of $\alpha$ by some $\prod t_i^{b_i}$. Having different $a_i$-s and $b_i$-s means that the rows and columns vary at different rates as I vary $t$, unless (a tentative conjecture) the rows or columns of $\alpha$ are unit vectors, or zero. Does that sound close to right? $\endgroup$ – Gyu Eun Lee Jun 8 '14 at 4:14
  • $\begingroup$ Your representations $\mathbb C_{A}$ are simple. Do you know about Schur's lemma? $\endgroup$ – Mariano Suárez-Álvarez Jun 8 '14 at 5:53

Here is my answer to my question. Thanks to Mariano Suarez-Alvarez for reminding me about Schur's lemma.

First, let us identify $U = \mathbb{C}^m$, $V = \mathbb{C}^{n+m}$, $W = \mathbb{C}^n$.

The decomposition of $\rho$ into irreducible representations $\rho = \mathbb{C}_{b_1}\oplus\cdots\mathbb{C}_{b_{n+m}}$ produces a decomposition of $\mathbb{C}$ into $n+m$ copies of $\mathbb{C}$. Consider the restriction of $\beta$ to one of those factors: we have the diagram $$ \require{AMScd} \begin{CD} \mathbb{C} @>{\beta}>> \beta(\mathbb{C}) \\ @V{\mathbb{C}_{b_i}}VV @VV{\psi_t\mid_{\beta(\mathbb{C})}}V \\ \mathbb{C} @>{\beta}>> \beta(\mathbb{C}) \end{CD} $$ Since $\beta(\mathbb{C})$ has dimension at most $1$, the restriction of $\psi_t$ is either a zero-dimensional or one-dimensional representation. The zero-dimensional representation is uninteresting, so suppose the restriction is a one-dimensional representation; the decomposition of representations into irreducibles implies that this is an irreducible representation. By Schur's lemma, $\beta$ restricted to this direct summand $\mathbb{C}$ must be zero or an isomorphism. Since $\beta$ is onto, this implies that $\beta$ sends $n$ summands injectively into $W = \mathbb{C}^n$, and the remaining $m$ summands are exactly $\ker\beta = \text{im}~\alpha$.

Consider the restriction of $\alpha$ to its direct summands as before. $\alpha$ is an isomorphism onto its image, so this produces isomorphisms of irreducible representations $$ \require{AMScd} \begin{CD} \mathbb{C} @>{\alpha}>> \alpha(\mathbb{C}) \\ @V{\mathbb{C}_{a_j}}VV @VV{\rho_t\mid_{\alpha(\mathbb{C})}}V \\ \mathbb{C} @>{\alpha}>> \alpha(\mathbb{C}) \end{CD} $$ So the representation $(V,\rho)$ breaks up into representations $V = \text{im}~\alpha\oplus \beta^{-1}(\mathbb{C}^{n}) = \mathbb{C}^m\oplus\mathbb{C}^n$, and $\rho = \sigma\oplus\pi$, where $\sigma$ is regarded as a representation $\text{im}~\alpha\to\text{im}~\alpha$ and $\pi$ as a representation $\beta^{-1}(\mathbb{C}^n)\to\beta^{-1}(\mathbb{C}^n)$; this is possible because we have shown that $\sigma$ and $\pi$ are isomorphic to such representations via the above diagrams.

This is also obviously a sufficient condition, so we conclude that a short exact sequence of representations $0\to\sigma\to\rho\to\pi\to 0$ exists if and only if $\rho \cong\sigma\oplus\pi$.


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