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A principal $G$-bundle over a space $X$ is classified by the homotopy classes of maps $[X,BG]$, where $BG$ is the classifying space of the group $G$. My question is what can we do about this when the fiber is a homogeneous space.

For a general fiber bundle $F\hookrightarrow E\rightarrow X$, the classification is usually done by $[X,B\rm{Diff}(F)]$ and is obviously a difficult beast. But for a homogeneous space $G/H$, for $G$ a Lie group and $H$ a closed subgroup, can we say something about the space $B\rm{Diff}(G/H)$? Is there any way to relate them to say $BG$ and $BH$?

More specifically what I'm looking for is a way to classify Lagrangian Grassmann bundles over a certain manifold. It is known that the fiber $LG(n,2n)=U(n)/O(n)$ is a homogeneous space. So is there any classification result in this direction?

Any help is appreciated!

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Unfortunately, just because $G/H$ a homogeneous space doesn't mean we get much control over $B\text{Diff}(G/H)$. At most you get something nice if $G/H$ is dimension $\leq 3$.

What you might have better luck with is by thinking of the Lagrangian Grassmannian as a homogeneous space: it has an effective action of $U(n)$, which means an injective homomorphism $f: U(n) \to \text{Diff}(U(n)/O(n))$.

You could restrict your transition functions to lie in the image of $f$. (One says you've "reduced the structure group to $U(n)$".) In this case, classifying Lagrangian Grassmannian bundles with structure group $U(n)$ is the same thing as classifying principal $U(n)$-bundles over your space: given a principal $U(n)$-bundle $P$ over a base $B$ (so it has a free right action of $U(n)$ whose quotient is $B$), we may take $P \times_{U(n)} \left(U(n)/O(n)\right)$ to get a Lagrangian Grassmannian bundle over $B$.

(Alternatively, look at the transition functions: they're exactly the same for the two bundles, the fibers are just different!)

So you've then reduced yourself to calculating $[X, BU(n)]$, aka, finding the set of rank $n$ complex vector bundles over your space.

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  • $\begingroup$ That’s actually a nice approach. Curious : what happens when the dimension is less than 3? $\endgroup$ – ChesterX Oct 5 '18 at 15:43
  • $\begingroup$ @ChesterX Diffeomorphism groups are better understood then. $U(1)/O(1)$ is just $\Bbb{RP}^1$, and has $S^1 \simeq \text{Diff}^+(\Bbb{RP}^1)$. I was going to write more about the next case, but then realized I didn't know how to describe $U(2)/O(2)$ very well. I think it's a non-orientable manifold of dimension 3 whose oriented double cover is $S^2 \times S^1$. I suspect the diffeomorphism group of that is actually calculated. Hatcher proved, for instance, that $\text{Diff}(S^2 \times S^1) \simeq \Omega SO(3) \times O(3) \times O(2)$. $\endgroup$ – user98602 Oct 5 '18 at 18:00
  • $\begingroup$ Decompose $\mathfrak{u}(2)$ as $\mathfrak{u}(2) = \mathfrak{o}(2)\oplus \mathfrak{p}$, orthogona with respect to a bi-invariant metric on $U(2)$. Then $O(2)$ acts on $\mathfrak{p}$ via the adjoint action. A basis for $\mathfrak{p}$ is given by the elements $\begin{bmatrix} 0 & i \\ i & 0\end{bmatrix}$, $diag(i,0)$, and $diag(0,i)$. The element $diag(1,-1)\in O(2)$ conjugates the latter two basis elements to themselves, but maps the first to its negative. This is orientation reversing, so $U(2)/O(2)$ is non-orientable. Since $U(2)/SO(2)$ is $S^1\times S^2$, your guess is correct. $\endgroup$ – Jason DeVito Oct 5 '18 at 20:55
  • $\begingroup$ @JasonDeVito Nice, thanks. What is $\pi_1(U(2)/O(2))$? This was the fundamental obstruction to me figuring out this space. I guess it is the unique nontrivial $S^2$-bundle over $S^1$. $\endgroup$ – user98602 Oct 5 '18 at 21:05
  • $\begingroup$ We have the diffeo $U(2)\rightarrow SU(2)\times S^1$ given by $A\mapsto (A', \det(A))$ where $A'$ has the same second row as $A$, but the first row of $A'$ is $\det(A)^{-1}$ times the first row of $A$. Under this diffeomorphism, the $SU(2)\subseteq O(2)$ doesn't change the determinant, so only acts on $SU(2)$, so $U(2)/SO(2) \cong S^2\times S^1$. Now, the element $diag(1,-1)\in O(2)$ acts on $S^2\times S^1$, we see that it acts as $-1$ on the $S^1$ factor, and it does something on the $S^2$ factor... $\endgroup$ – Jason DeVito Oct 5 '18 at 21:28

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