Are these vector bundles, trivial bundle?

Question

We identify the vector space tensor product $\mathbb{R}^{m} \otimes \mathbb{R}^{n}$ with $\mathbb{R}^{mn}$

Let $X$ be the space of all non zero simple tensors $X=\{a\otimes b \mid a\in \mathbb{R}^{n} \setminus \{0\}, \;b\in \mathbb{R}^{m} \setminus \{0\}\}$. Let $\pi:\mathbb{R}^{mn}\setminus \{0\} \to \mathbb{R}P^{(mn-1)} $ be the natural projection. Put $PX=\pi (X)$

Is the tautological line bundle restricted to $PX$, a trivial bundle? Is the following bundle $(E,X,q))$ a trivial bundle over $X$:

$E=\{(x\otimes y, T) \mid T:E_{x} \to E_{y} \;\; \text{is a linear map }$ where $E_{x}= \{v\in \mathbb{R}^{n} \mid v.x=0\} $

$X$ is the space simple tensors and $q$ is the obvious projection.

Answer 1

There is an evident map $\mathbb{R}P^{m-1}\times\mathbb{R}P^{n-1}\to PX$, which is easily seen to be an isomorphism. (This is called the Segre embedding.) If we pull back the tautological bundle of $PX$, we get the tensor product of the tautological bundle $M$ over $\mathbb{R}P^{m-1}$ and the tautological bundle $N$ over $\mathbb{R}P^{n-1}$. Let $x$ and $y$ denote the first Stiefel-Whitney classes of $M$ and $N$, so the mod $2$ cohomology of $PX$ is $\mathbb{F}_2[x,y]/(x^m,y^n)$. The first Stiefel-Whitney class of $M\otimes N$ is $x+y\neq 0$, so $M\otimes N$ is not trivial.

The rest of your question is notationally unclear. I will interpret it as follows: you want to consider the bundle $M^\perp$ over $\mathbb{R}P^{m-1}$, whose fibre at $\mathbb{R}x$ is the orthogonal complement to $x$. You also want to consider the bundle $N^\perp$ over $\mathbb{R}P^{n-1}$, and the resulting bundle $E=\text{Hom}(M^\perp,N^\perp)$ over $PX$. (You said $X$ but I am guessing that you meant $PX$. If you pull back to $X$ then $E$ certainly becomes stably trivial, but not obviously trivial. Different methods would be needed to address that.) Now $M^\perp$ has an obvious inner product, so it is isomorphic to its dual, so $E$ can be identified with $M^\perp\otimes N^\perp$. Using $M\oplus M^\perp=m$ and $N\oplus N^\perp=n$ we obtain an isomorphism $$ E \oplus (m\otimes N) \oplus (M\otimes n) \simeq mn \oplus (M\otimes N). $$ Each of these bundles $V$ has a total Stiefel-Whitney polynomial $f_V(t)=\sum_iw_i(V)t^{\text{dim}(V)-i}$, and standard methods give \begin{align*} f_{M\otimes n}(t) &= (t + x)^n \\ f_{m\otimes N}(t) &= (t + y)^m \\ f_{M\otimes N}(t) &= t+x+y \end{align*} The above isomorphism therefore gives $$ f_E(t) (t+y)^m (t+x)^n = t^{mn} (t+x+y). $$ From this you can calculate $f_E(t)$ and thus $w_i(E)$ for all $i$. The details depend on the parity of various binomial coefficients which appear when you expand out $(t+x)^n$ and $(t+y)^m$. However, you will always get $w_1(E)=nx+my+(x+y)=(n+1)x+(m+1)y$, so $E$ can only be trivial if $n$ and $m$ are odd. However, in that case it works out that $w_2(E)$ always contains $xy$ (and possibly also $x^2$ and/or $y^2$), so again $E$ is nontrivial.