What is an example of a short exact sequence of finite dimensional $ \mathbb{C}[G] $ modules
$$
0 \to W \to V \to \mathbb{C} \to 0
$$
which does not split?
In other words, what is an example of a finite dimensional complex representation $ (G,V) $ that surjects $ G $-equivariantly onto the trivial representation $ \mathbb{C} $ but has no trivial subrepresentation?
Note that if $ G $ is a linearly reductive group (for example any finite group, compact group, semisimple group, or the complex points of a compact group e.g. the general linear group) then every $ G $ representation is completely reducible so the SES must split.
Pf. Since $ G $ is linearly reductive then $ V $ and $ W $ are completely reducible so $ V= \bigoplus_\rho m_\rho \rho $ and $ W= \bigoplus_\rho n_\rho \rho $ with $ \rho $ all irreducible. Then
$$
V/W= \bigoplus_\rho (m_\rho-n_\rho) \rho \cong \mathbb{C}
$$
so we must have $ m_\rho=n_\rho $ for all nontrivial $ \rho $ and $ m_\rho=1+n_\rho $ when $ \rho $ is the trivial irrep. Let $ 1 $ denote the trivial irrep. Let $ W_1 $ denote the $ 1 $ isotypic subspace of $ W $ and let $ V_1 $ denote the $ 1 $ isotypic subspace of $ V $. Recall $ m_1=n_1+1 $ so $ W_1 $ has codimension $ 1 $ in $ V_1 $. Since $ G $ is linearly reductive we can take a complement of $ W_1 $ in $ V_1 $, call it $ U $
$$
V_1 = W_1 \oplus U
$$
The map $ k \to U $ is the desired splitting.
What I am asking about is groups $ G $ that are not linearly reductive so this argument fails. Any counterexample is ok, but an example with $ G $ a complex linear algebraic group and the short exact sequence algebraic is best.
I thought about some basic nonreductive group like the solvable affine group $ \begin{bmatrix} a & b \\ 0 & \frac{1}{a} \end{bmatrix} $ or the nilpotent Heisenberg group $ \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} $ but I couldn't get any obvious representations to work as counterexamples.
Best Answer
Example of finite-dimensional $V$: Let $V = \mathbb{C}^2$ and define: $$G = \left\{ \begin{bmatrix} a & b \\ 0 & 1\end{bmatrix} : a \in \mathbb{C}^{\times}, b \in \mathbb{C}\right\}$$ It is easy to see that $V^G = 0$ by direct computation. Moreover, the $1$-dimensional subspace $W$ spanned by $(1, 0)$ is a $\mathbb{C}G$-submodule and $V/W$ is trivial (as seen using $W^{\perp}$).
Example of infinite-dimensional $V$: Let $G = \mathrm{Aut}(S)$ for a countably infinite set $S$ and $V$ be the vector space $\oplus_S k$ with the standard $G$-action. Let $W$ be the codimension $1$ subspace with sum of coordinates equal to zero. We have $V/W$ being trivial, but $V^G = 0$.