Geometric interpretation of reduction of structure group to $SU(n)$. 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.
 A: 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 $E$.
