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Let $\xi$ be a (real) vector bundle of dimension $n$. Then the first Stiefel-Whitney class $$ w_1(\xi)=0 $$ if and only if $\xi$ is orientable, i.e. the structure group of $\xi$ can be reduced to $SO(n)$.

Question 1: Let $\xi^\mathbb{C}$ be a complex vector bundle of dimension $n$. Then the first Chern class $$ c_1(\xi^\mathbb{C})=0 $$ if and only if the structure group of $\xi^\mathbb{C}$ can be reduced to $SU(n)$? Is it true or false? Any references?

Question 2: Let $\xi^\mathbb{H}$ be a quaternion vector bundle of dimension $n$. Then the first Pontrjagin class $$ p_1(\xi^\mathbb{C})=0 $$ if and only if the structure group of $\xi^\mathbb{H}$ can be reduced to $SSp(\mathbb{H},n):=\{A\in Sp(n)\mid \text{Det}(A)=1\}$? Is it true or false? Any references?

I have a further question characteristic classes of a covering space with symmetric group action

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    $\begingroup$ "the structure group" isn't well-defined, unless you put a metric on it and ask about the holonomy group, in which case the answers are "false". You can ask whether the structure group can be reduced to $SU(n,\mathbb C$ or $\mathbb H)$, in which case the answers are "true". This is standard characteristic class stuff, as in Milnor's book or Husemoller's, not research-level. $\endgroup$
    – Allen Knutson
    Commented Oct 6, 2015 at 2:06
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    $\begingroup$ Crossposted from MSE. $\endgroup$
    – mme
    Commented Oct 6, 2015 at 4:18
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    $\begingroup$ Could you please explain your definition $SU(\mathbb H,n)$? The notation is nonstandard. I did not find it on the wikipedia page in the link. And to relate to Sullivan's answer: quaternion vector bundle means vector bundle with an $\mathbb H$ action by scalars, doesn't it? So the group $Sp(n)\hat\times Sp(1)$ will not appear in this context. $\endgroup$
    – Sebastian Goette
    Commented Oct 6, 2015 at 6:27
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    $\begingroup$ It it probably worth pointing out that there is no nontrivial normal subgroup $\mathrm{SSp}(\mathbb{H},n)\subset \mathrm{Sp}(n)$ (other than the center, which is discrete) because the group $\mathrm{Sp}(n)$ is simple. Thus, there is no interpretation of $p_1(\xi)=0$ as a condition for reducing the structure group of a quaternionic vector bundle $\xi$ to such a subgroup. $\endgroup$
    – Robert Bryant
    Commented Oct 6, 2015 at 11:41
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    $\begingroup$ Regardless of this particular question, in my opinion, "standard characteristic class stuff" like "standard applications of spectral sequences" like "standard facts about Floer homology" etc. are research-level mathematics as far as mathoverflow is concerned. $\endgroup$
    – Gil Kalai
    Commented Oct 6, 2015 at 12:50

2 Answers 2

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The natural question also relates to understanding holonomy in Riemannian geometry using the idea of $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and Cayley numbers as scalars. There are 8 discussions, two for each of the four choices of scalars, whether non orientable or orientable in each context.

Orientable : $SO(n)$, $SU(n)$, $Sp(n)$, and $G_2$ where $G_2$ is the 14-dimensional exceptional Lie group and non-orientables are group structures on twisted products of the four orientables with the unit spheres in each space of scalars. Namely $O(n)$, $U(n)$, $Sp(n)\cdot Sp(1)$, and $Spin(7)$. This is a current topic of research, discussing geometries that realize the groups in this exhaustive list of holonomy groups in Riemannian geometry which are maximal, not symmetric space holonomy groups and not products.

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I think haohaizi was looking for the following standard facts:

$BSO(n)$ is the fiber of $Bdet : BO(n) \longrightarrow BO(1)$ and $Bdet = w_1$. This uses $BO(1) = K(Z/2,1)$.

$BSU(n)$ is the fiber of $Bdet : BU(n) \longrightarrow BU(1)$ and $Bdet = c_1$. This uses $BU(1) = K(Z,2)$.

There is no determinant function for quaternions; they are not commutative. As Sullivan suggests, $Sp(n)$ is already `special', i.e., oriented. Further, $Sp(1) = S^3$ is not a $K(Z,3)$, so the obstruction to orientability here is not a cohomology class.

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