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$([x],v)\mapsto [x,x.v]$):


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?

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    $\begingroup$ You clearly cannot have a one-point remainder. Because if you had, then you would obtain $G(2,n+1)$ from $G(2,n)$ by gluing in a $2(n-1)$-cell. Inductively, $G(2,n)$ would have the same homology as $\mathbb C P^{n-2}$, but already $G(2,3)\cong\mathbb R P^2\not\cong\mathbb C P^1$. $\endgroup$ Oct 21, 2015 at 6:32
  • $\begingroup$ @SebastianGoette thank you very much for your very interesting comment. For such cell-gluing, are you considering the Thom space of the bundle? $\endgroup$ Oct 21, 2015 at 7:26
  • $\begingroup$ Moreover are we sure that these gluing maps induce the same maps as the $\mathbb{C}P^{n},s$( in relative homologies $H_{n}(X_{n+1}, X_{n})$)? $\endgroup$ Oct 21, 2015 at 7:32
  • $\begingroup$ could you please more explain about cell structure and gluing cells? $\endgroup$ Oct 21, 2015 at 7:58
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    $\begingroup$ Regarding the cell structure obtained by gluing - it is not clear that you get the same as for $\mathbb C P^n$. But that does not matter if you are interested in homology only, because the resulting CW complex would only have cells in even dimensions. Then the cellular complex would have trivial differential, and hence be the same as for $\mathbb C P^n$ with the usual cell structure. $\endgroup$ Oct 21, 2015 at 8:46

3 Answers 3


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.

  • $\begingroup$ why $G(k,n+1)$ is non orientable for $1<k,n$? $\endgroup$ Oct 22, 2015 at 20:59
  • $\begingroup$ Maybe, $G(k,n+1)$ is orientable, but the tautological $\mathbb R^k$-bundle is not, so there is a nontrivial Stiefel-Whitney class. I edited the answer accordingly. $\endgroup$ Oct 23, 2015 at 8:18

There's two kinds of $k$-planes in $\mathbb{R}^n\times \mathbb{R}$: Those that project isomorphically to $\mathbb{R}^n$, and those that contain $\mathbb{R}$.

The former are given as graphs of linear functions $V\rightarrow \mathbb{R}$, where $V\subset \mathbb{R}^n$ is a $k$-plane. With the canonical scalar product on $\mathbb{R}^n\times\mathbb{R}$ restricted to $V$, linear functions on $V$ naturally correspond to vectors. This means that we just constructed a bijection between tuples $(V,v), V\subset \mathbb{R}^n, v\in V$ and an open subset of $Gr(k, n+1)$. The former is the usual description of the tautological bundle on $Gr(k, n)$.

The complement (the planes that intersect $\mathbb{R}$) is in natural bijection to $Gr(k-1,n)$, so we can see that that is your complement. In particular, you'll only get a point for $k=1$, the $\mathbb{R}P^n$-case.

  • $\begingroup$ Thank you for very interesting answer. Can one imagine some other embedding which remainder is a compact manifold with a different topological type (different from $Gr(k-1,n)$) $\endgroup$ Oct 22, 2015 at 21:27
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    $\begingroup$ Alexander-Poincare duality (as in Hatcher, Algebraic Topology, Thm. 3.44) shows that for any embedding of $Gr(k,n)$ into $Gr(k,n+1)$, the cohomology of the complement is related by a long exact sequence to the homology of $Gr(k,n)$ and the cohomology of $Gr(k,n+1)$. This suggests that the topology of the complement is very tightly constrained. To prove that there's no other possible complement is probably very hard, though. Still, this excludes the point and many other spaces essentially simpler than $Gr(k-1,n)$. $\endgroup$ Oct 23, 2015 at 17:37

$\newcommand{\bRP}{\mathbb{RP}}$ $\newcommand{\bR}{\mathbb{R}}$ The result about the complement of a point in $\bRP^{n+1}$ is related to the natural cell decomposition of $\bRP^{n+1}$. The counterpart of this decomposition for higher Grassmannians is the so called Schubert decomposition and you can find a particularly readable description in Chapter 6 of Milnor & Stasheff's classic Characteristic Classes.

These cell decompositions have a Morse theoretic description, and this point of view will enable you to construct embeddings of many homogeneous spaces in to Grassmannians.

Fix an $n$-dimensional dimensional Euclidean space $V$ and denote by $\DeclareMathOperator{\Gr}{\boldsymbol{Gr}}$ $\Gr_k(V)$ the Grassmanian of $k$-dimensional subspaces of $V$. $\DeclareMathOperator{\Sym}{Sym}$. For a subspace $S\in\Gr_k(V)$, denote by $P_S$ the orthogonal projection onto $S$ viewed as a symmetric operator $P_S: V\to V$. Denote by $\Sym(V)$ the space of symmetric linear operators $V\to V$.

The correspondence

$$\Gr_k(V)\ni S\mapsto P_S\in\Sym(V) $$

produces a smooth embedding $\Gr_k(V\hookrightarrow \Sym(V)$.

The space $\Sym(V)$ is equipped with a natural inner product $\DeclareMathOperator{\tr}{tr}$

$$(A,B)=\tr(AB),\;\;\forall A,B\in \Sym(V). $$

This induces a Riemann metric on $\Gr_k(V)$.

Any operator $ A\in \Sym(V)$ defines a linear function $\ell_A:\Sym(V)\to\bR$, $B\mapsto \tr(AB)=(A,B)$. We denote by $f_A$ the restriction of $\ell_A$ to $\Gr_k(V)$.

For generic $A$ the function $f_A:\Gr_k(V)\to \bR$ is a Morse function. We denote by $\nabla f_A$ the gradient of $f_A$ with respect to the induced metric and by $\Phi_A^t$ the flow on $\Gr_k(V)$ generated by $-\nabla f_A$. Assuming $A$ generic, i.e., it has distinct eigenvalues, then the unstable manifolds of this flow are precisely the Schubert cells giving the Schubert cellular decomposition described by Milnor and Stasheff.

When $A$ is not generic $f_A$ is not necessarily Morse but it is Morse-Bott. In this case the critical submanifolds of $f_A$ are intersting homogeneous spaces. For example, if you take $A$ to be the orthogonal projection on a $1$-dimensional subspace $L$,then the restriction of $\ell_A$ to $\Gr_1(V)$ is Morse-Bott. Its absolute minima form a critical submanifold diffeomorphic to $\Gr_1(L^\perp)$, where $L^\perp$ is the orthogonal complement of $L$ in $V$. This function has a unique maximum, the point $L\in\Gr_1(V)$. From these two facts you get the statement about the complement of a point in $\Gr_1(V)$ mentioned at the begining of your question.

One can use the same function

$$\Gr_2(V)\ni S\mapsto \tr(P_LP_S)\in\bR $$

to obtain other interesting embeddings. For more details and other examples see this very nice article by Dynnikov and Veselov and Chapter 3 of my book on Morse theory.

Update 1. Here is an answer to your question. The Grassmannian $\Gr_2(\bR^{n-1})$ embeds in $\Gr_2(\bR^n)$. Using the above notation observe that $\Gr_2(L^\perp)$ ($2$-planes in $L^\perp$) embeds in $\Gr_2(V)$. The normal bundle of this embedding is the tautological $2$-plane bundle over $\Gr_2(L^\perp)$. This submanifold consists of the minima of the function $f_A$ where $A=P_L$. The complement of tubular neighborhood is not a disk though.

The maxima of the function $f_A$ consists of $2$-planes containing $L$. It is not hard to see that this set can be identified with lines in $V$ perpendicular to $L$, i.e., $\Gr_1(L^\perp)$. The normal bundle of this embedding is quotient tautological bundle, i.e.,the quotient of the trivial bundle $$ L^\perp\times \Gr_1(L^\perp)\to\Gr_1(L^\perp)$$ by the universal line bundle over $\Gr_1(L^\perp)$. Since the critical points of the function $f_A$ are either global minima or global maxima we deduce shows that $\newcommand{\bD}{\mathbb{D}}$

$$\Gr_2(\bR^n)= \bD_{\Gr_2(\bR^{n-1})}\cup_\partial \bD_{\Gr_{n-2}(\bR^{n-1})}, $$

where $\bD_{\Gr_k(V)}$ denotes the unit disk bundle of the tautological vector bundle over $\Gr_k(V)$, and $\cup_\partial$ denotes the gluing of two manifolds along their diffeomorphic boundaries.

  • $\begingroup$ Thank you very much for your interesting answer and your book on Morse theory. I was reading the first pages of your book: a question on Hessian: is it equivalent the geometric definition:$X.\Delta_{Y}\nabla f $ with respect to a metric and its connection? $\endgroup$ Oct 26, 2015 at 16:55
  • $\begingroup$ I think that you can do this computation yourself. $\endgroup$ Oct 26, 2015 at 20:37

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