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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)

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2 Answers 2

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.

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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.

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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
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@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
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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

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