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Related to this question, suppose we define $G(k,n)$ to be the set of $n\times k$ matricies with rank $k$, equipped with the quotient topology of $\mathbb{R}^{nk}$ by the equivalence relaiton $$A\sim B\iff \exists g\in GL(k,\mathbb{R}), A = Bg.$$

To show $G(k,n)$ is compact, we only need to show that $F(k,n)$ is compact, where $F(k,n)$ is the set of $n\times k$ matrices with rank $k$. As a subset of $\mathbb{R}^{kn}$, we need show that $F(k,n)$ is closed and bounded in Mat$_{n\times k}$.

When I try to show closedness, I observe that $A\in F(k,n)$ iff there exists a $(k,k)$-minor that has nonvanishing determinant. Thus I realized $F(k,n)$ has a finite union of complements of zero sets of polynomials, which is open. This is to the opposite of what I was expecting. Please help.

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mez
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    This approach isn't going to work, as the set in question is most definitely unbounded. Try thinking about Gram-Schmidt and reducing to the case of an orthonormal basis. – Ted Shifrin Aug 27 '14 at 23:28

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Consider the Stiefel manifold $V_k(\mathbb{R}^n)$ of orthonormal $v_1, \dots, v_k\in \mathbb{R}^n$. It's closed and bounded in $\mathbb{R}^{nk}$ and thus compact. Identifying points of $G(n, k)$ with $k$-dimensional subspaces of $\mathbb{R}^n$ gives a map $V_k(\mathbb{R}^n) \to G(n, k)$ that's sufficient to prove the compactness of the latter. The details are given in, for example, Section 5.1 of Milnor and Stasheff.

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