We know that for a complex vector bundle $E$ and curvature $F_{\nabla}$ on it.Then we can define the top Chern class as $det(\frac{i}{2\pi}F_{\nabla})$.
Then we view $E$ as a even dimension real bundle and define a Euler class on it via the Pfaffian.
My question is, do these two class equal with each other(maybe differ by a constant)?(not just represent the same cohomology class)
Thank you for your answer!
Best Answer
$\require{amsfonts} \require{amsmath}$ Let $A\in \mathfrak{u}(n)$ and let $A_{\mathbb{R}} \in \mathfrak{so}(2n)$ be the associated real matrix. Then one has $$ \det(iA) = Pf(A_\mathbb{R}).$$ If $n=1$, $A= i\lambda$ for $\lambda \in \mathbb{R}$ then $A_\mathbb{R} = \begin{bmatrix} 0 & \lambda \\ -\lambda & 0 \end{bmatrix}.$ You can calculate $$Pf(A_\mathbb{R}) = -\lambda = \det(iA). $$ In general this follows by diagonalizing $A$, then $A_\mathbb{R}$ has $2\times 2$ blocks on its diagonal and for any $2n\times 2n$ matrix the Pfaffian transforms like this: $$ Pf(B A_\mathbb{R} B^\top) = \det(B)Pf(A).$$
Edit: I wanted to add that there are two different sign conventions here. Since the Pfaffian is usually normalised requiring that $$ Pf\left(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\right) = \pm 1$$ (either plus or minus $1$) one might has to adjust the definitions. In Milnor & Stasheff's book they use the $+1$ convention and define the total chern class by $$ c(A) = \det(I + A/(2\pi i)).$$ While for example wikipedia uses $+1$ and $c(A) = \det(I + iA/(2\pi))$ (see here), so they should define $e(A_\mathbb{R}) = Pf(-A/(2\pi))$.