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This question is extensively rewritten by David Speyer; the original version is below.

The Grassmannian $G(k,n)$ is the quotient $SU(n)/S[U(k) \times U(n-k)]$. Let's write $\pi$ for the map $SU(n) \to G(k,n)$.

Suppose that I have an $k(n-k)-1$ form $\omega$ on $SU(n)$ which is invariant under the $S[U(k) \times U(n-k)]$ action. Then the restriction of $\omega$ to each $S[U(k) \times U(n-k)]$ orbit will be closed. So $d \omega$ will restrict to $0$ on each orbit, and will be $S[U(k) \times U(n-k)]$ invariant. This implies that there is a differential form $\eta$ on $G(k,n)$ so that $d \omega = \pi^{\ast} \eta$. (Actually, I'm not sure whether this implies it in general. But it happens in the cases that the OP gives in comments below, and in any case we can assume that there is an $\eta$ with $d \omega = \pi^{\ast} \eta$.) We can abuse notation and refer to $\eta$ as $d \omega$.

Is there a variant of Stokes theorem which will allow us to compute $\int_{G(k,n)} \eta$ by integrating $\omega$ over some $k(n-k)-1$ cycle? What is this $k(n-k)-1$ cycle?


What is the compact manifold that we regard as the boundary of complex Grassmanians when applying Stokes theorem to the integral of a 2-form over the Grassmanian?

In case of $\mathbb{CP}^k$, for instance, this becomes $S^{2k-1}$ so that $$\int_{\mathbb{CP}^k} F = \int_{S^{2k-1}} A$$ where $F:=dA$.

Thanks in advance AB

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    $\begingroup$ I cannot make sense of yr formula: $\mathbb{CP}^k$ has no boundary (neither do any of the Grassmannians) so when I integrate an exact $2k$-form $F$ over it, I get zero. $\endgroup$ – Fran Burstall Jul 18 '13 at 21:37
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    $\begingroup$ You can onlyintegrate a $2$-form over a two-dimensional manifold. Grassmannians generally donot have dimension $2$. $\endgroup$ – Mariano Suárez-Álvarez Jul 18 '13 at 21:48
  • $\begingroup$ @ Fran Well not generically! Take the vector potential $A=-iu^{*}_jdu_j$ (Cartan-Maurer form) on the sphere where $u_i$ are spinor coordinates satisfying $u^{*}_ju_j=1$. Then using stereographic map one can show that $\int_{\mathbb{CP}^1} dA = 2\pi$ which gives Dirac quantization rule. Indeed this is due to the bundle structure of the complex projective space ${\mathbb{CP}^1}=S^2=SU(2)/U(1)$. $\endgroup$ – Alireza Behtash Jul 18 '13 at 22:04
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    $\begingroup$ I am not sure I understand yr notation but I think you are integrating the curvature of a connection on a non-trivial line-bundle over $\mathbb{CP}^1$. Now locally, this connection (in a trivialisation) looks like $d+A$ and then the curvature is $dA$ but this is a local expression only: yr vector potential is not globally defined as a $1$-form on $\mathbb{CP}^1$. $\endgroup$ – Fran Burstall Jul 18 '13 at 22:46
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    $\begingroup$ I'm pretty sure I know what the question is asking. I'm going to take the liberty of editing this question to what I think is being asked. $\endgroup$ – David E Speyer Jul 19 '13 at 13:16
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wat?

(sorry, I can't make comments yet. It seems you're trying to integrate a bundle-valued differential form, rather than ordinary one. Maybe you want a(n orthonormal) Stiefel manifold? http://en.wikipedia.org/wiki/Stiefel_manifold)

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  • $\begingroup$ You cannot define a non-vanishing abelian vector potential $A$ on a Stiefel manifold because $U(1)$ gauge does not exist in this space. Look at its coset representation. $\endgroup$ – Alireza Behtash Jul 18 '13 at 22:19
  • $\begingroup$ sorry, I'm not sure what you just wrote means. I'm afraid you need someone who speaks better physics than me. (what is an abelian vector potential?) $\endgroup$ – John Salvatierrez Jul 18 '13 at 22:26
  • $\begingroup$ I'm sorry for not being clear. It is basically the Cartan-Maurer form created from the $U(1)$ generator of the $SU(n)$ group via $A=inTr(t^lg^{-1}dg)$ where $t^l$ stands for the $U(1)$ generator of the $SU(n)$ in $Gr(k,n)= SU(n)/ SU(n-k)\times SU(k)\times U(1)$. $n$ is the so-called $U(1)$ charge which is zero in the case of Stiefel manifolds. Here $g\in SU(n)$. $\endgroup$ – Alireza Behtash Jul 18 '13 at 22:37

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