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Edit: I remove the trivial part of the first version, according to comment of Alex Degtyarev

Let $M$ be the manifold of all matrices in $M_{n}(\mathbb{R})$ with fixed rank $0<k<n$. There is a canonical $n-k$ bundle $E$ over $M$. $E$ is the kernel bundle:$$\{(A,v)\in M_{n}(\mathbb{R})\times \mathbb{R}^{n}\mid Av=0\}$$

We consider the $Z_{2}$ action on base space $M$ via transpose operation

Question: Can one say that: This action can not be lifted to a $\mathbb{Z}_{2}$- equivariant structure for $E\to M$?

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  • $\begingroup$ $\ker(GA)=\ker A$ $\endgroup$ Mar 4, 2015 at 9:20
  • $\begingroup$ @AlexDegtyarev Yes I was wrong. I revise the question. $\endgroup$ Mar 4, 2015 at 9:23
  • $\begingroup$ Definitely, one can say this. $\endgroup$
    – Sasha
    Mar 4, 2015 at 11:30
  • $\begingroup$ @Sasha Why? are you considering the action of $Z_{2}$ on characteristic class? Could you please more explain? $\endgroup$ Mar 4, 2015 at 11:41
  • $\begingroup$ @AlexDegtyarev According to your comment answer to my simillar question, there is a circle in the manifold for which the bundle loose orientability(thanks again for your very elegance intuition) :$$\begin{pmatrix}cos^{2}t&costsint\\costsint&sin^{2}t\end{pmatrix}$$. But the $Z_2$ transpose action is fix(trivial) here. So the restricted bundle is obviously equivariant. But what about globally? $\endgroup$ Mar 5, 2015 at 14:06

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The transposition action does not lift, at least, "typically". The easiest obstruction is in the case $k=1$; in general, I think it's the same, with a bit more characteristic classes (or, "projective space" replaced with other magic words like "Grassmannian"). So, restrict the bundle to matrices with a fixed image $V=\Bbb R\subset\Bbb R^n$. The bundle in question is essentially the cotautological bundle over the "almost" projective space $\Bbb R^n\smallsetminus0=\operatorname{Hom}(\Bbb R^n,\Bbb R)\smallsetminus0$. But the action takes this subset to matrices whose kernel is the fixed space $V^\perp$, to which the restriction of the bundle is trivial.

Added: In fact, the action obviously lifts if $k=0$ or $n$, and does not lift otherwise. For proof, as above, one can stick to the projective space: just restrict the bundle to the set of block-diagonals of the identity $\operatorname{id}\colon\Bbb R^{k-1}\to\Bbb R^{k-1}$ and rank $1$ matrices $\Bbb R^{n-k+1}\to V\subset\Bbb R^{n-k+1}$ with fixed image $V$.

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  • $\begingroup$ I think that I can not understand something in your argument at least when n is even. Ex:n=4 matrix with fixed image $\mathbb{R}$ are in the form $$\begin{pmatrix}a&b&c&d\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$. So the bundle is both trivial on this set and on its transpose. The same example works for all even n. $\endgroup$ Mar 9, 2015 at 15:28
  • $\begingroup$ I correct "At least for n=2,4,8" $\endgroup$ Mar 9, 2015 at 15:37
  • $\begingroup$ And for nontriviality of cotautological bundle we do not have this restriction $n\neq 2,4,8$. So is not a contradiction here?Am I missing something? $\endgroup$ Mar 9, 2015 at 15:40
  • $\begingroup$ I see: you do not projectivize this time, do you. Then, you are right: coranks $1$, $3$, $7$ will need special treatment. I doubt that they are really special (as that was just a subset), but who knows? $\endgroup$ Mar 9, 2015 at 16:05
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    $\begingroup$ Say, for corank $1$, it suffices to consider $n=2$, $k=1$. Then the map sending an operator to its kernel and image maps your space to $S^1\times S^1$ (with contractible fibers), and $w_1$ of your bundle is the generator of $H_1$ of the first factor. On the other hand, the transposition interchanges the factors, thus rendering a non-isomorphic bundle. $\endgroup$ Mar 9, 2015 at 16:10

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