# The Grassmanian as a Homogenous Space and Related Spaces

I am interested in studying a quotient of Lie groups related to the Grassmanian. I don't know very much topology so this question will be a little bit open ended.

Let $p \neq q$ and consider the complex Grassmanian: $SU(p+q) / S(U(p)\times U(q))$. I am interested in $SU(p+q) / SU(p)\times SU(q))$. This is a spherical space in the sense of http://archive.numdam.org/ARCHIVE/CM/CM_1979__38_2/CM_1979__38_2_129_0/CM_1979__38_2_129_0.pdf

It is a bigger space than the Grassmanian (since the quotient is smaller).

Is there a theory that would help me understand the spherical space because of this relationship? Is there a way to transfer properties from the well studied Grassmanian to this space? For general homogenous spaces, do we have theorems relating such 'intermediate' spaces? I also welcome any comments from the perspective of Lie algebras.

I hope this is not too vague.

-

(I don't speak German, so I don't know what the paper you linked to is about. The following is mostly about the topology of your example, which I'll call $X$.)

Consider a chain of closed subgroups $H\subseteq K\subseteq G$. This gives rise to a so called homogeneous fiber bundle $$K/H\rightarrow G/H\rightarrow G/K.$$

The projection map sends $gH$ to $gK$. Further, in the case that $H$ is normal in $K$, this fiber bundle is in fact a principal $K/H$ bundle. The element $kH$ acts on $gH$ via $kH *gH = gk^{-1}H$.

This is well defined: if $k = k'h_1$ and $g = g'h_2$, then $$kH * gH = gk^{-1}H = g'h_2h_1^{-1}k'^{-1}H = g'h_3 k'^{-1} H = g'k'^{-1} h_3' H = g'k'^{-1}H.$$ The element $h_3$ is simply, by definition, $h_2h_1^{-1}$ and the element $h_3'$ is defined by $k'h_3k'^{-1} = h_3'$, using normality of $H$ in $K$.

It is also an action as $kk'H*gH = gk'^{-1}k^{-1}H = kH*(k'H* gH)$. Finally, it's free since if $k'H gH = gH$, then $k'^{-1}\in H$ to $k'\in H$ so $k'H = H$.

Applying this to your example with $H = SU(p)\times SU(q)$, $K = S(U(p)\times U(q)$, and $G = SU(p+q)$, and noting that $K/H = S^1$ (as Lie groups) via the map sending an element of $S(U(p)\times U(q))$ to the determinant of the $U(p)$ part, we see that $X$ is a principal $S^1$ bundle over the Grassmanian, $Gr$.

In fact, more can be said. Principal $S^1$ bundles are classified by homotopy classes of maps from the Grassmanian into $BS^1 = \mathbb{C}P^\infty = K(\mathbb{Z},1)$. So, canonically, these are classified by $H^2(Gr)$.

It's not too hard to show from the LES in homotopy groups coming from the fibration $S(U(p)\times U(q))\rightarrow SU(p+q)\rightarrow Gr$ that $Gr$ is simply connected with $\pi_2(Gr) = \mathbb{Z}$. It follows from the Hurewicz theorem that $H^2(Gr) = \mathbb{Z}$. (There is also a well known cellular decomposition which gives you this).

Using a similar LES in homotopy groups for $X$, you can see that $H^2(X)$ is trivial. (Further, you can easily that $pi_k(Gr)$ is isomoprhic to $\pi_k(X)$ for all $k\geq 3$.)

It follows that the Euler class of the the principal $S^1$ bundle is a generator of $H^2(Gr)$. This, paired with the Gysin sequence, should allow you to compute the cohomology of $X$ in terms of that of $Gr$.

Finally, if you want to know the characteristic classes of (the tangent bundle to) $X$, notice that $TX = \nu\oplus\pi^* TGr$ where $\nu$ is the trivial rank 1 bundle (coming from the $S^1$ orbits) and that, after using the Gysin sequence, you know what $\pi^*$ does on cohomology.

-
This is all very helpful, thank you. I should have expanded a little bit on the definition of spherical space in the article I linked: For $G$ compact, connected Lie group, and $H$ a closed subgroup, $H$ is a spherical subgroup if for every irreducible unitary representation $/rho$ of $G$, the fixed point set of $H$ is at most one dimensional. It is not so relevant to the question, and I thought I would include it for motivation. Would you mind recommending some sources to learn more about these specific kinds of bundles? –  Johannes Wachs Nov 4 '11 at 19:49
@Johannes: I actually don't know any sources that have this kind of info - I just kind of worked it out as I went along. Hopefully someone else can come along and provide some. Incidentally, there are other ways of directly computing the cohomlogy ring and characteristic classes. The only reference I have for these other ways is some hand written notes I got in grad school. –  Jason DeVito Nov 4 '11 at 22:55