MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As we all know, for any homogeneous space $G/H$ we have that the equivariant vector bundles over $G/H$ are characterized by the representations of $H$. Thus, for the the complex projective line $CP^1 \simeq SU(2)/U(1)$, it must hold that all its line bundles are indexed by the integers $L_k$, for $k \in Z$, and more generally, its rank-$k$ (equivarian) vector bundles are of the form $$ L_{\bf z} = L_{z_1} \oplus \cdots \oplus L_{z_k}, {\text ~~~ for ~~~ } {\bf z} \in Z^k. $$ Does this then extend to all the total flag manifolds $F(n)$, ie the spaces of the form $$ F(n) := SU(n)/(U(1)^{\otimes n-1}). $$

Edit: I omitted the word equivariant by mistake and have now entered it as (equivariant)

share|cite|improve this question
up vote 7 down vote accepted

I think perhaps the confusion stems from the following.

It is true that the category of $G$-equivariant $G$-bundles on $G/H$ is equivalent to the category of representations of $H$. The flag variety $\mathcal F l_n$ can be realized as either $U(n)/U(1)^n$, or as $GL_n(\mathbb C)/B_n$, where $B_n$ is the group of upper triangular matrices.

Thus $GL_n(\mathbb C)$-equivariant vector bundles on $\mathcal Fl_n$ are equivalent to representations of $B_n$. This group is not reductive, and not every representation is a direct sum of 1-dimensional representations. Thus, not every equivariant vector bundle is a direct sum of line bundles.

On the other hand, the category of $U(n)$-equivariant principal $U(n)$-bundles is equivalent to representations of $U(1)^n$. This category is semisimple.

share|cite|improve this answer
Thanks for your answer. Just one question: Are saying that all simple reps of $U(1)^n$ are 1-dimensional, and hence that my guess that all U(n)-equiv bundles are constructable from line bundles? – Ago Szekeres May 1 '13 at 13:51

The answer is no. For example, if $n = 3$ then $F(3)$ is a divisor of bidegree $(1,1)$ in $P^2\times P^2$ and the pullback of the tangent bundle from any factor is an example of an equivariant bundle which is not a sum of line bundles.

On the other hand, any equivariant bundle on $F(n)$ can be obtained as an iterated extension of line bundles.

share|cite|improve this answer
What is an iterated extension of line bundles? – Ago Szekeres Apr 30 '13 at 18:27
also, what is the representation of $U(1)^{\otimes 2}$ corresponding to the vector bundle you give as a counterexample? – Ago Szekeres Apr 30 '13 at 18:28
@Ago: An iterated extension is a $G$-equivariant vector bundle (or, equivalently, locally free sheaf) together with a $G$-invariant filtation by $G$-equivariant vector subbundles (locally free subsheaves with locally free quotient) whose associated subquotients are each $G$-equivariant line bundles (invertible sheaves). I am confused by your notation for $U(1) \times \dots \times U(1)$. The representation of this group is the same as the adjoint representation of this group on $\mathfrak{sl}_{3}/\mathfrak{b}$, where $\mathfrak{b}$ is upper triangular $3\times 3$ matrices with trace zero. – Jason Starr Apr 30 '13 at 18:41
Sam Gunningham's answer cleared this up, but another point view is: although vector bundles on $SL_n/B$ may not split algebraically, on the real manifold $F(n) = SU(n)/U(1)^{n-1}$, every extension of vector bundles does split: choose a hermitian metric. (Indeed, identifying $F(n)$ this way is essentially doing just that.) – Dave Anderson Apr 30 '13 at 23:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.