I learned from the following post that the total space of the tautological line bundle over $\mathbb{R}P^{n}$ is diffeomorphic to $\mathbb{R}P^{n+1}\setminus \{pt\}$.(There is a natural embedding$([x],v)\mapsto [x,x.v]$):


<https://math.stackexchange.com/questions/1486105/is-the-total-space-of-the-tautological-line-bundle-over-mathbbrpn-a-non>


Now assume that $E$ is the total space of the tautological 2-plane bundle over the real Grassmanian $G(2,n)$.


As a generalization of the above fact we ask:



> 
> Is there a (natural) embedding of $E$ into $G(2,n+1)$? If yes, what is the topological type of the remainder $G(2,n+1)\setminus E$. Is it possible to have an embedding with a one point remainder?
> 
> 
> 



A natural embedding of the total space of tautological bundle over $G(2,n)$ in $G(2,n+1)$
To answer the final question, assume that the total space of the taulogical bundle $E\to G(k,n)$ embeds into $G(k,n+1)$ such that $G(k,n+1)\setminus E$ consists of a single point only. Then $G(k,n+1)$ is the Thom space of $E$. The Thom isomorphism for $E$ shows that $H^\ell(G(k,n+1);\mathbb Z/2)=0$ for $0<\ell<k$.


On the other hand,
the Grassmannian $G(k,n+1)$ approximates the classifying space $BGL(k,\mathbb R)$, and the map $G(k,n+1)\to BGL(k,\mathbb R)$ is $(n+1)$-connected. Hence, there is always a nontrivial homomorphism $\pi\_1(G(k,n+1))\to\pi\_1(BGL(k,\mathbb R))\cong\mathbb Z/2$, so $H^1(G(k,n+1);\mathbb Z/2)\ne 0$. But this implies that for $k\ge 2$, no embedding as above exists.


