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Let $E \to X$ be a complex vector bundle of rank $k$. Then the structure group of $E$ can be reduced to $U(k)$, as this is equivalent to specifying a hermitian inner product on $E$ which can always be done using a partition of unity (or more high-browly, since $U(k)$ is homotopy equivalent to $GL(k,\mathbb C))$. My question is if there is a nice geometric structure on $E$ that is equivalent to having a reduction of the structure group to $SU(k)$.

Note that I am not talking about the holonomy group but the group that the transition functions take values in.

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The possibility of reducing the structure group from $U(n)$ to $SU(n)$ has geometric interpretation, but not in terms of extra structure of the bundle but as a property of the bundle (which I think is something different).

So let me explain this in detail. Let $E \to X$ be a complex vectorbundle of rank $n$ and $X$ be a reasonable space. Then we know $E$ is classified by some map $\varphi: X \to BU(n)$. The question about the reduction of the structure group then means we want to find a lift of $\varphi$ to $SU(n)$, i.e., we want a lift (up to homotopy) in the diagram

$$\begin{array} = & & BSU(n) \\ && \downarrow \\ X & \stackrel{\varphi}{\longrightarrow} & BU(n) \end{array}. $$

Now consider the fiber sequence $$ SU(n) \to U(n) \stackrel{det}{\to} U(1),$$ this implies that applying the classifying space functor yields a fibration $$ BSU(n) \to BU(n) \stackrel{Bdet}{\to} BU(1).$$

This in turn implies that the functor $\lbrack X, -\rbrack$ of homotopy classes of maps out of $X$ yields an exact sequence of the form

$$ \lbrack X, BSU(n) \rbrack \to \lbrack X, BU(n) \rbrack \to \lbrack X,BU(1) \rbrack $$ which tells us that we can lift the map $\varphi$ to $BSU(n)$ if and only if the composite map $$X \stackrel{\varphi}{\to} BU(n) \stackrel{Bdet}{\to} BU(1) $$ is nullhomotopic. But here comes the geometric interpretation into play, because of the following fact: Consider the functor $det$ from Vectorbundles to (line)bundles. If $E$ is a vectorbundle over $X$ of rank $n$ then the bundle $det(E)$ has as fiber at $x\in X$ the $n$th exterior power $\Lambda^n(E_x)$. By abstract nonsense (i.e. the Yoneda Lemma) we know that this functor is represented by a map $BU(n) \to BU(1)$ and it turns out this map is precisely $Bdet$. This tells us that there is a reduction of the structure group of $E$ to $SU(n)$ if and only if the determinant line bundle $det(E)$ is trivial.

This should also not be too surprising because in terms of the transition functions in local trivializations the determinant bundle $det(E)$ arises by taking the transition functions of $E$ and applying det to them. But if you want these transition functions to have values in $SU(n)$ then after applying det all transition functions become multiplication by $1$, hence $det(E)$ is trivial.

But let me continue just one more step and give a precise statemet of when this reduction is possible in terms of characteristic classes.

Recall that there is an isomorphism $c_1: LineBundles(X) \to H^2(X,\mathbb{Z})$ which is given by taking the first chern class of line bundles in integral cohomology. Hence we know that the possibility of reducing the structure group of $E$ to $SU(n)$ is given if and only if $c_1(det(E)) = 0$. But it is also a fact that in integral cohomology there is the equation $c_1(E) = c_1(det(E))$ and hence reducing the structure group of $E$ to $SU(n)$ is possible if and only if $c_1(E) = 0$ in $H^2(X,\mathbb{Z})$, and the geometric interpretation is given by triviality of the determinant line bundle $det(E)$ of $E$.

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Very very nice answer. I wish I knew more about classifying spaces, can you recommend somewhere to read about them? –  Olivier Bégassat May 1 '12 at 10:28
Thanks :) actually I am not really sure where to read about this stuff. But I suppose in Hatchers Book or in Bredon's Geometry and Topology there should definitely be some information. I think as soon as you know what they are it is very helpful to start to translate geometrical constructions you know from vectorbundles to homotopy theory (via the maps that classify the bundles). For me it was a lot of "learning by doing". –  mland May 1 '12 at 10:40
One of the standard references is Milnor & Stasheff. It's a very pleasant read. –  Aaron Mazel-Gee May 1 '12 at 11:27
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