# A line bundle over the manifold of singular matrices

According to answer of Denis Serre to this question, the manifold of singular matrices in $M_{n}(\mathbb{R})$ is defined as follows: $$M=\{A\in M_{n}(\mathbb{R})\mid \text{rank}(A)=n-1\}$$

So we define a (line bundle) over this manifold: $$\{(A,v)\in M\times\mathbb{R}^{n}\mid Av=0\}$$.

Is it a trivial line bundle?

No, this bundle is not trivial (starting from dimension $2$). Introduce a metric, consider projector to a hyperplane, and rotate this hyperplane through $\pi$ about an axis. You get an orientation reversing loop.

• Thank you very much for your answer.Before that I post this question, I tried a lot to introduce a nonvanishing global section(for $n=2$). I could not find a global section. But during the computation I realize that I can not prove the local trivialization around $2\times 2$ matrices with one zero row. So befor that I learn your answer, could you help me for this local trivialization. On the other hand the restriction of this bundle to matrices without zero rows is trivial. Is not this fact contradictory to your answer?I mean does not your method work for such trestriced bundle? Feb 18, 2015 at 19:59
• What is the problem? I'm not good with formulas, but just take the orthogonal projection of the kernel of you variable matrix to the kernel of the fixed one. For a while, it's an isomorphism. Feb 18, 2015 at 20:10
• BTW, I didn't think much about this, but I don't see why the restriction to matrices without zero rows should be trivial: a whole row of zeroes has too high a codimension to affect $w_1$. Feb 18, 2015 at 20:12
• OK, if $n=2$, having a zero row has codimension $1$, so $w_1$ may change. Concerning my construction, in this dimension, you cannot rotate a line without hitting a coordinate line. Feb 18, 2015 at 20:28
• Block-diagonal of identity and the $2$-dimansional $\mathbf x\mapsto\langle\mathbf x,\mathbf a\rangle\mathbf a$, where $\mathbf a(t)=(\cos t,\sin t)$, $t\in[0,\pi]$. Feb 19, 2015 at 13:51

Consider the natural map $M \to\mathbb RP^n$ assigning $x\mapsto ker x$. Then obviously your line bundle arises as the pullback of the tautological bundle. But now you can write down a homotopically non-trivial map $S^1 \to M \to \mathbb RP^n$ (try to restrict your attention to hit $\mathbb RP^1\subset \mathbb RP^n$). No it is a matter of taste to say that the Stiefel Whitney class will be non trivial or considering the non-trivial pullback to $S^1$ etc.

• Actually, you can do even better: $\mathbb{R}P^n$ is a retract of $M$. Just pick a metric and consider orthogonal projections. Feb 22, 2015 at 5:18
• Note that $M$ consists of matrices of rank $n-1$, not of all singular matrices.
– abx
Feb 22, 2015 at 10:59
• @Dan +1 for your interesting answer. I am sorry that I can not accept two answers. Feb 23, 2015 at 16:15
• @Dan Can one repeat your argument for the complex case? Feb 23, 2015 at 16:21
• you will still get a map to $\mathbb CP^n$. Then the line bundle is trivial if and only if this map is null homotopic. Feb 24, 2015 at 2:24