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Being more familiar with linear operators than with geometry, I like to see $G_K(r,n)$, the Grassmannian of all $r$ dimensional subspaces in $K^{r+n}$ ($K=\mathbb{R}, \mathbb{C}$), as the set of all rank $k$ projections (self-adjoint idempotents) in $M_n(K)$. And I am also interested in the infinite-dimensional Grassmannians sitting in $B(H)$, the algebra of bounded linear operators on a separable infinite-dimensional $K$ Hilbert space. In the latter, the projections split into connected components according to their rank and their nullity. These are smooth manifolds, but modelled on an infinite-dimensional (for $r\neq 0$ and $n\neq 0$) Banach space. They are the infinite-dimensional analogues of the above. I cheated with the usual notations to include the latter: for instance, $G_\mathbb{C}(3,\infty)$ corresponds to the rank $3$ projections in $B(H)$.

Thinking about some linear questions, I ended up being interested in the following: does there exist a nowhere vanishing section of the tangent bundle of $G_K(r,n)$? The two cases I am really comfortable with are $G_\mathbb{R}(1,1)$ (yes, that's $S^1$ and the tangent bundle is trivial), and $G_\mathbb{C}(1,1)$ (no, by the hairy ball theorem since it is $S^2$).

I vaguely know that under some conditions, the tangent bundle of a manifold admits a nowhere vanishing section if and only if its Euler class (or should say number?) is zero. And that under some conditions, this coincides with the Euler characteristic of the manifold.


1 - What is a precise statement and set of conditions for the latter, hopefully applying to the (possibly infinite-dimensional) Grassmannians?

2 - If it applies, what are the relevant Euler invariants (characteristic/class/number) for $G_K(r,n)$? After a lot of googling, I think I found $\binom{r+n}{r}$ in the complex case (without being sure because I am having a hard time understanding the geometric language). Is this true? What does the number mean exactly? What about the real case? What about the infinite-dimensional case?

3 - Ultimately, I am mostly interested in knowing when I can say that any section of the tangent bundle of $G_K(r,n)$ must vanish, and how to justify it properly. Could you clarify this for me?

4 - I think it is about time for me to understand these things. Do you know a friendly reference for someone who has a lot of difficulty with geometry?

Thank you.

Edit: forget the infinite-dimensional analogue. In $B(H)$, the unitary group is much less twisted than in $M_n$. A famous theorem of Kuiper shows that it is contractible. If I am not mistaken, this entails that the Grassmannians of $B(H)$ have a trivial tangent bundle.

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Is that a "protesting" nickname? =) – Pedro Tamaroff Aug 5 '13 at 21:53
up vote 6 down vote accepted

1 and 2) The precise statement is usually given in the Poincare-Hopf index theorem. The computation of the Euler characteristic involves the binomial coefficients, yes. And I think you have the answer in the complex case correct. Sometimes in the Poincare-Hopf index theorem they fail to mention that if the euler characteristic is zero (on a connected manifold), you can find a vector field with no zeros. But the proof in Milnor and Stasheff's Characteristic classes book shows you how to do it.

3) I put the computation of the Euler characteristic of the Grassmann manifolds in the Wikipedia page (look at the Schubert cell section)

4) This work is not algebraic geometry. It is differential topology. It's written up well in Guillemin and Pollack. Also in Bredon's Algebraic Topology text

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Thank you very much. Is this the correct statement? It does not even state the other direction: that if there is a nowhere vanishing section, the Euler characteristic is $0$. Is it implicit in that the sum on the left is zero if void? Also, are there any known generalizations in the infinite-dimensional case? – 1015 Aug 1 '13 at 23:50
Yes, sums over an empty set are conventionally taken to be zero. What infinite-dimensional case are you referring to, could you be precise? – Ryan Budney Aug 1 '13 at 23:51
Ok, thanks. Since the complex Grassmannians are orientable, and the real ones are orientable iff $n$ is even, Poincare-Hopf answers my question in the finite-dimensional case, except for the real case in odd dimension. That's great, thanks again. – 1015 Aug 2 '13 at 0:32
I'm confused why you make the distinction orientable/non-orientable. If the Euler characteristic is zero, the manifold has an everywhere non-zero vector field, even if the manifold is non-orientable. – Ryan Budney Aug 16 '13 at 4:10

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