When does a short exact sequence of representations exist? 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.
 A: 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 $\rho \cong\sigma\oplus\pi$.
