# Example of map to the trivial representation that does not split

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.

• This is only a counterexample if we allow infinite-dimensional $V$: let $S$ be a countably infinite set, $G$ be the group of bijections of $S$, $V$ be the vector space $\oplus_S k$ with the standard $G$-action, and $W$ be the submodule with sum of coordinates equal to zero. We have $V/W$ being trivial, but the corresponding short exact sequence does not split since $V^G = 0$. Commented Jan 15 at 1:08
• Perhaps one can take $V = \mathbb{C}^2$ and $G$ to be the stabilizer of some $1$-dimensional subspace $W$ inside $\mathrm{GL}(V)$? This cannot be split (otherwise, $V$ would be the direct sum of trivial representations, which is not the case). Commented Jan 15 at 1:12
• You are asking two different questions, but both have negative answer in the sense that there are examples. Commented Jan 15 at 1:35

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 $$V^G = 0$$.
• an even simpler example is take $a=1$ in your example so $G \cong (\mathbb{C},+)$. See for example here math.stackexchange.com/questions/4821105/… Commented Jan 16 at 22:26