Short Exact Sequence Vector Spaces

Question

I've been posed the above question, but I'm a bit unclear what it's asking. Is $N$ any normal subgroup of $GL(U,V)$? The homomorphism $\phi: GL(U,V) \to GL(U) \times GL(V/U)$ I plan on using is (I think?) the natural one $g \mapsto (g|U,\tilde{g})$ where $\tilde{g}: V/U \to V/U$ is the map $\tilde{g}(xU) = g(x)U$.

So my questions are (1): What is $N$? (2): Will this be the right homomorphism to use?

EDIT: also how to think about the isomorphism $N \simeq \text{Hom}(V/U,U)$? I understand intuitively why this is true I think, since if $g \in N$, $\phi(g) = (1_U,1_{V/U})$ which says $g(V/U) \subset U$ and $g(U) = U$. But what confuses me is how to construct an isomorphism where on one side we have linear map composition and on the other we have addition?

Answer 1

In the finite dimensional case you can think in terms of matrices:

Choose a basis $u_1,...,u_d,u_{d+1}...,u_n$ of ( V ), such that the first $d$ form a basis of $U$. In this basis, an element of $GL(U,V)$ is a block matrix

$$ \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} =: (A,B,C) $$

where the submatrix $A$ corresponds to the restriction to ( U ).

The multiplication law is $(A,B,C) \cdot (A',B',C') = (AA', AB'+BC',CC')$.

Your homomorphism $GL(U,V) \to GL(U)\times GL(V/U)$ is given by $ (A,B,C) \mapsto (A,C) $ and the kernel consists of elements of the form $ (I,B,I) $, where $B$ is any $n-d\times d$ matrix. This corresponds to a map $Hom(V/U,U)$, because the $n-d$ vectors $$ u_{d+1}+U,...,u_n+U $$ form a basis of $V/U$ and the $d$ vectors $$ u_1,...,u_d$$ were a basis of $U$.

Surely a more abstract answer would be more satisfying, however, this view may be useful for you as well :)

PS: Allow me to use the same shortcut as above: The group law for elements of the form $(I,B,I)$ is $(I,B,I)\cdot (I,B',I) = (I,B+B',I)$, which explains the addition.

Answer 2

Once you find a surjective homomorphism $$ \mathit{GL}(U,V)\to\mathit{GL}(U)\times\mathit{GL}(V/U) $$ the (normal) subgroup $N$ will be its kernel (and it has to be, if you want an exact sequence, so no choice allowed).

The homomorphism you are considering is the only natural choice (but elements of $V/U$ are usually written as $x+U$, as the operation is addition).

You just need to show it's surjective, besides being a homomorphism.

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The subgroup $N$ consists of all elements $g\in\mathit{GL}(U,V)$ such that $g|U$ and $\tilde{g}$ are the identity, that is \begin{align} &g(x)=x &&\text{for all $x\in U$} \tag{*}\\ &g(v)+U=v+U &&\text{for all $v\in V$} \tag{**} \end{align} How do you get a linear map $V/U\to U$ from these data? Attached to $v+U$, you have $g(v)-v\in U$, by (**). Such an element doesn't depend on the particular $v$; indeed, if $v+U=v'+U$, then $v-v'\in U$, so $$ (g(v)-v)-(g(v')-v')=g(v-v')-(v-v')=0 $$ because $g(v-v')=v-v'$ by (*). The map $\hat{g}\colon v+U\mapsto g(v)-v$ is clearly linear.

Now your task is to prove that the assignment $g\mapsto \hat{g}$ is the required (group) isomorphism, after having shown that $\hat{g}$ is indeed linear.