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I am recently learning representation theory and come across with the example that the additive group $\mathbb{R}$ acts on $V=\mathbb{R}^2$ by $\begin{pmatrix} 1&x\\ 0&1\end{pmatrix} \begin{pmatrix} v_1\\ v_2\end{pmatrix}= \begin{pmatrix} v_1+xv_2\\ v_2\end{pmatrix}$

I know that this representation is reducible because it has a proper subspace $W=\{(w,0)^T|w\in \mathbb{R}\}$ which the group acts on. But it is indecomposable: Indecomposable but not irreducible representation and direct sums

I understand the proof in the above link but I would like to ask isn't the representation decomposable because we have the decomposition $V\cong W \oplus V/W$ and both $W$ and $V/W$ are non trivial?

Thank you so much for your help.

Ishigami
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    $V$ is not isomorphic to $W\oplus V/W$. Why did you think it would be? – Jyrki Lahtonen Apr 16 '22 at 12:34
  • @JyrkiLahtonen I am sorry I thought it is true that for every finite dimensional vector space $$ with $$ a vector subspace, we have $V\cong U \oplus V/U$ – Ishigami Apr 16 '22 at 12:53
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    Only isomorphic as vector spaces, here we discuss isomorphic as representations. The group $\Bbb{R}$ acts trivially on the direct sum $W\oplus V/W$. That is, each $x\in\Bbb{R}$ acts via the identity mapping, but that is not the case with $x$ acting on $V$. Hence they are not isomorphic. – Jyrki Lahtonen Apr 16 '22 at 13:49
  • @JyrkiLahtonen Thank you so much. – Ishigami Apr 16 '22 at 14:05

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