Let $X$ be a compact Riemann surface of genus $g$, then $K^1_{\mathrm{top}}(X)\cong\mathbb{Z}^{2g}$. Is there a explicit description of a set of basis of $K^1_{\mathrm{top}}$? (e.g., For cohomology $H^1(X,\mathbb{Z})\cong\mathbb{Z}^{2g}$ we may take the 1-cochains “around the holes'')
Furthermore, we define the Mukai vector of $\kappa\in K^1_{top}(X)$ to be $v(\kappa)=\mathrm{ch}(\kappa)\sqrt{\mathrm{td}(X)}$, and the Euler pairing on $K^*_{top}(X)$ by $\langle a,b\rangle=(v(a^\vee),v(b))$ where $(-,-)$ is the pairing in $H^*(X,\mathbb{Q})$. Do we know the pairing with respect to the basis? (The definition of $\mathrm{ch}\colon K^*(X)\to \oplus H^*(X)$ is given in https://www.maths.ed.ac.uk/~v1ranick/papers/ahvbh.pdf)
Best Answer
Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.
The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^{2g} = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)
Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq (\bigvee_{2g} S^2) \vee S^3$, which gives $[\Sigma M^g, BU] \cong \pi_2 (BU)^{2g} \oplus \pi_3 (BU) = \mathbb{Z}^{2g}$ (and by Bott periodicity $[\Sigma M^g, BU] \cong [M^g, \Omega BU] \cong [M^g, U]$). See Tyrone's comment below for a way to make this explicit.