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Let $V_{n}(\mathbb{C}^{k})$ the Stiefel manifold of $n$-frame in $\mathbb{C}^{k}$. We can see $V_{n}(\mathbb{C}^{k})$ as a subset of $n$ copies of the cartesian product $S^{2k-1} \times \cdots \times S^{2K-1}$. So we have a bundle $ V_{n-1}(\mathbb{C}^{k-1}) \rightarrow V_{n}(\mathbb{C}^{k}) \rightarrow S^{2k-1} $.

How can I find local trivializations?

Chris Brooks
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ArthurStuart
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    My first thought is general theory: Given compact Lie groups $H\subseteq K\subseteq G$, there is always a fiber bundle of the sort $K/H\rightarrow G/H\rightarrow G/K$. In this case, $G = U(k)$, $K = U(k-1)$ and $H = U(k-n)$ where $U(N)$ is the complex unitary group of $N\times N$ matrices. I'm too busy now to expand this into a full answer (if this is even the kind of answer you want), but in several hours I'll have some time. – Jason DeVito - on hiatus Jan 23 '13 at 13:49
  • You'll very helpfull if you give me later a full answer. Thanks – ArthurStuart Jan 23 '13 at 14:02
  • But $U(k)/U(k-1)\simeq S^{2k+1}$... – ArthurStuart Jan 23 '13 at 15:03
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    Actually, $U(k)/U(k-1) = S^{2n-1}$. Consider for example $U(1)/U(0) = S^1 = S^{2(1)-1)} \neq S^{2(1)+1}. I'll start writing up my answer now. Not exactly sure how long it will take or how much detail to go into. – Jason DeVito - on hiatus Jan 23 '13 at 18:08
  • Your «So we have a bundle...» claim does not follow from what is before it. – Mariano Suárez-Álvarez Jan 23 '13 at 20:14

1 Answers1

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As I mentioned in the comments, given compact groups $H\subseteq K\subseteq G$, there is always a fiber bundle of the form $$K/H\rightarrow G/H \rightarrow G/K$$ where the projection map $\pi$ maps $gH$ to $gK$ and which has structure group $K$.

I don't know of any particularly easy or obvious proof of this fact. The only reference I have for it is a bunch of hand written notes of my advisor, but there are probably other references out there. The key lemma is that one always has a slice and that the isotropy representation in such a case must be trivial.

The most general form of it I know is: if $K$ is a compact Lie group acting smoothly on a manifold $M$ with all isotropy groups conjugate (say, to a subgroup $H$), then the projection $M\rightarrow M/K$ is a fiber bundle map with structure group $K$ and fiber $K/H$.

Anyway, using this on the inclusions $U(k-n)\subseteq U(k-1)\subseteq U(k)$ (via the usual block embeddings) gives you the fibration you're looking for.

For your particular example, if $G$ acts transitively on a manifold $M$ with isotropy subgroup $H$, then there is a $G$-equivariant diffeomorphism $G/H\cong M$. This is proven in Lee's Introduction to Smooth Manifolds Theorem 9.24.

So, in order to show, for example, $U(k)/U(k-1)\cong S^{2k-1}$, it suffices to find a transitive $U(k)$-action on $S^{2k-1}$ with isotropy subgroup $U(k-1)$. Well, $U(k)$ naturally acts on $\mathbb{C}^k$ and preserves lengths of vectors, so it also acts on $S^{2k-1}\subseteq\mathbb{C}^k$. It's not too hard to see this action is transitive. The set of elements in $U(k)$ which fix the North pole is easily seen to be $U(k-1)$ block embedded.

For the Stiefel manfolds, we have $U(k)/U(k-n)\cong V_n(\mathbb{C}^k)$. Note that $U(k)$ naturally acts on $\mathbb{C}^k$, and hence on the collections of ordered tuples of vectors in $\mathbb{C}^k$. Given any $n$ orthonormal vectors $\{v_1,..., v_n\}$, complete them to an orthonormal basis $\{v_1,..., v_n, w_{n+1}, ... ,w_k\}$. The matrix whose columns are the $v$'s and $w$'s is an element of $U(k)$ which maps the first $n$ standard basis vectors to $\{v_1,...,v_n\}$.

This proves that $U(k)$ acts transitively on $V_n(\mathbb{C}^k)$. I'll leave it to you to check that the matrices in $U(k)$ which fix the first $n$ standard basis vectors are precisely the matrices of the form $\operatorname{diag}(I,B)$ with $B\in U(k-n)$.

Jason DeVito - on hiatus
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  • Assuming you are the same ArthurStewart, this question: http://math.stackexchange.com/questions/284510/covering-space-weyl-group-flag-manifold is also answered by this method as a covering space is nothing but a fiber bundle with discrete fiber. – Jason DeVito - on hiatus Jan 23 '13 at 19:54
  • I should also add that, for example, one doesn't need $H$, $K$, and $G$ to be compact, but rather each should be closed in the next. I typically work solely with compact Lie groups, so I often forget when hypothesis can be weakened to, say, proper actions of noncompact groups, etc. – Jason DeVito - on hiatus Jan 23 '13 at 20:00
  • I don't know how we can answer to my question on covering spaces with you answer to this post. – ArthurStuart Jan 23 '13 at 21:17
  • @ArthurStuart: Use $H = T$, $K = N_G(T)$, $G = G$. (Also, if you want your accounts merged, you can flag something for moderator attention and ask them to do it.) – Jason DeVito - on hiatus Jan 23 '13 at 21:43
  • But tha action of $U(n)$ on $V_n(\mathbb{C}^k)$ is free? Right? But how works? – ArthurStuart Jan 24 '13 at 09:06
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    Did you swap an $n$ and a $k$? The group $U(n)$ acts on $V_n(\mathbb{C}^k)$ by multiplying the first $n$ coordinates of each vector, but this action doesn't have to be free. (Consider a pair of vectors in $\mathbb{R}^4$ with $0$s in the first two coordinates). There's also an action of $U(k)$ on $V_n(\mathbb{C}^k)$, but this action is not free - it's transitive. – Jason DeVito - on hiatus Jan 24 '13 at 14:11
  • But the action of $U(n)$ on $V_{n}(\mathbb{C}^k)$ must be free, because $V_n(\mathbb{C}^{k})/U(n)$ is the Grassmannian manifold! So the action hasn't fixed points... – ArthurStuart Jan 24 '13 at 14:35
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    I see your point. There are multiple actions of $U(n)$ on $V_n(\mathbb{C}^k)$ and I picked the wrong one. The correct (free) action of $U(n)$ on an $n$-tuple of $k$-vectors given as follows. Let $p=(v_1,..., v_n)\in V_n(\mathbb{C}^k)$ and consider $p$ as a $k\times n$ matrix. Then $U(n)$ can multiply on the right and this action is free. – Jason DeVito - on hiatus Jan 24 '13 at 14:49
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    And why is free? I think is obvious... – ArthurStuart Jan 24 '13 at 15:37
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    It's pretty obvious. Suppose $A$ fixes $p$. Since the columns of $p$ are orthogonal, they are independent, so in particular, the row rank is $n$. That is, the rows of $p$ span $\mathbb{R}^n$. The action of $A$ on $p$ acts on each row and multiplies one at a time, so if $A$ fixes $p$, then it fixes each row, so it fixes their span = $\mathbb{R}^n$, so $A = I$. – Jason DeVito - on hiatus Jan 24 '13 at 16:16
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    Sorry... then also $G \rightarrow G/N_{G}(T)$ is a principal $N_{G}(T)$-budle? – ArthurStuart Jan 24 '13 at 18:53
  • I'm am quite confused: $A$ and $\mathbb{R}^{n}$... – ArthurStuart Jan 24 '13 at 19:08
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    $A$ is an element of $U(n)$. When it multiplies $p$ (on the right), it really just multiplies each row of $p$ one at a time. If $A$ fixes $p$, then it fixes each row. Now, interpret the rows as points in $\mathbb{R}^n$. Since the row rank is $n$, the row vectors of $p$ span $\mathbb{R}^n$. Since $A$ fixes each row vector, it fixes any linear combination of them, hence fixes this entire $\mathbb{R}^n$. This means that $wA = w$ for all $w\in\mathbb{R}^n$ which is exactly what it means for $A$ to be equal to $I$. And yes, $G\rightarrow G/N_G(T)$ is a principal $N_G(T)$ bundle. – Jason DeVito - on hiatus Jan 24 '13 at 19:40
  • And the proof of the fact that $G \rightarrow G/N(T)$ is a principle $N(T)$-bundle is the same? – ArthurStuart Jan 24 '13 at 21:57
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    It's the general statement that if $H$ acts on $M$ freely, then $M\rightarrow M/H$ is an $H$-principal bundle I alluded to. I don't know of a particularly easy proof of this fact. – Jason DeVito - on hiatus Jan 25 '13 at 00:03
  • I posted your first answer in an other web and they (under assunction that $G \rightarrow G/K$ is a principal bundle) gave me a simple proof: http://mathoverflow.net/questions/119751/lie-groups-bundle/119794#119794 However I don't understand how he builds the isomorfism and the associate bundle... – ArthurStuart Jan 25 '13 at 00:27
  • I had actually seen that earlier :-). I look at both sites pretty regularly, but don't really contribute on the other one much. – Jason DeVito - on hiatus Jan 25 '13 at 00:29
  • But do you think that the proof is right? I don't understand how he builds the isomorphism and the associate bundle... – ArthurStuart Jan 25 '13 at 00:46
  • I confess I didn't read it in detail. I'm actually kind of tired tonight, so I won't be doing any more math. But I'll definitely take a look at it tomorrow. Sorry! – Jason DeVito - on hiatus Jan 25 '13 at 00:48
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    I finally got around to reading his proof and it looks fine to me. Perhaps you could ask another question (here) on exactly what's confusing you. (In fact, I kind of few the calculations as a good exercise in understanding bundle isomorphisms, etc). – Jason DeVito - on hiatus Jan 25 '13 at 18:17
  • I just saw that you did exactly that earlier. In that case, I would follow Martin's suggestion and expand it into a full fledged question. – Jason DeVito - on hiatus Jan 25 '13 at 18:36
  • shouldn't the structure group be K/H instead of K? – Mr. Brown Apr 05 '19 at 19:31
  • @Paul T: First, unless $H$ is normal in $K$, $K/H$ typically doesn't have a group structure. Second, I did mean $K$. The point is you have a $K$-prinicpal bundle $G\rightarrow G/K$ (with $K$ acting by right multiplication by inverses). From the left $K$-action on $K/H$ (by left multiplication), you can form the associated bundle $(K/H\times _K G)\rightarrow G/K$ which has fiber $K/H$. It's not too hard to prove that $K/H\times_K G$ is diffeomorphic to $G/H$, so you've associated the bundle $K/H\rightarrow G/H\rightarrow G/K$ to a principal $K$-bundle. So $K$ is the structure group. – Jason DeVito - on hiatus Apr 06 '19 at 19:42
  • @JasonDeVito yeah sorry i mixed up fiber and structure group. very new to the material – Mr. Brown Apr 06 '19 at 20:00