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Let $V$ be a finite-dimensional vector space over a field $K$. Consider the group $GL(V)$ of non-singular linear maps acting on pairs of subspaces $(U,W)$ of fixed dimensions $p$, $q$ respectively by $g(U,W)=(gU,gW)$. Prove that two pairs $(U,W)$ and $(U',W')$ are in the same $GL(V)$-orbit (i.e., there exists $g\in GL(V)$ such that $gU=U'$ and $gW=W'$) $\Leftrightarrow \dim(U\cap W)=\dim (U'\cap W')$.

This came up on a qualifying exam, and frankly, and I don't even know where to begin in order to relate the dimension to orbits. I thought maybe an appeal to the index of the space might help, but that was fruitless. Any suggestions?

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3 Answers 3

up vote 2 down vote accepted

Note that $g(U\cap W)\subseteq g(U)\cap g(W)$. If $gU=U'$ and $gW=W'$, then this implies that $\dim(U\cap W)\leq \dim(U'\cap W')$. Applying the same argument with $g^{-1}$ we obtain $\dim(U'\cap W')\leq \dim(U\cap W)$, proving equality.

For the converse, assume that $\dim(U\cap W) = \dim(U'\cap W') = r$.

Fix a basis $x_1,\ldots,x_r$ for $U\cap W$, and a basis $y_1,\ldots,y_r$ for $U'\cap W'$. Let $u_{r+1},\ldots,u_{p}$ be such that $\{x_1,\ldots,x_r,u_{r+1},\ldots,u_p\}$ is a basis for $U$, and let $w_{r+1},\ldots,w_{q}$ be such that $\{x_1,\ldots,x_r,w_{r+1},\ldots,w_{q}\}$ is a basis for $W$. Note that $\{x_1,\ldots,x_r,u_{r+1},\ldots,u_p,w_{r+1},\ldots,w_q\}$ is linearly independent (prove it!), so we can complete it to a basis $\beta$ for $V$.

Similarly, pick $u'_{r+1},\ldots,u'_p$ so that $y_1,\ldots,y_r,u'_{r+1},\ldots,u'_p$ is a basis for $U'$, $w'_{r+1},\ldots,w'_{q}$ to complete the $y$s to a basis for $W$, and complete $\{y_1,\ldots,y_r,u'_{r+1},\ldots,u'_p,w'_{r+1},\ldots,w'_q\}$ to a basis $\gamma$ for $V$. Then define $g\in \mathrm{GL}(V)$ by mapping $x_i$ to $y_i$, $u_j$ to $u'_j$, $w_k$ to $w'_k$, and the rest of the basis $\beta$ to what remains in $\gamma$.

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Which direction do you find trouble with?

Implication $\Leftarrow$ can be shown by explicitly constructing the automorphism: choose any basis $B_\cap$ of $U\cap W$, extend it to a basis $B_U$ of $U$, and a basis $B_W$ of $W$. Then $B_U\cup B_W$ is a basis of $U+W$. Extend it to a basis $C$ of X. In the same way, choose $B_\cap',B_W',B_U',C'$. Let $f\in GL(V)$ be such that $f[B_\cap]=B_\cap'$ etc. It can be chosen because $C$ and $C'$ are bases, and it clearly maps $(U,W)$ to $(U',W')$.

$\Rightarrow$ is easy: if $f(U,W)=(U',W')$, then $f(U\cap W)=U'\cap W'$ (since $f$, as a bijection, preserves set-theoretic operations), and $f$ as an automorphism of $V$ preserves dimensions of subspaces, which immediately yields the result.

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For one direction, if $g$ exists then it is easy to show that about $g(U\cap W)\subset(U'\cap W')$ and $g^{-1}(U'\cap W')\subset(U\cap W)$. This would establish the dimensions are the same.

For the other direction, choose a basis for $(U\cap W)$ and extend to a basis of $V$ in way that allows you to define $g$ nicely.

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