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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)

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    $\begingroup$ There's a map K(Z,1)~U(1)--> U and I'd imagine that induces an isomorphism betwen H^1 and K^1 in this case. At the very least it works for n=1 (and n=0...) $\endgroup$
    – kiran
    Commented Mar 5, 2022 at 23:16
  • $\begingroup$ It would be helpful to explain exactly what is meant by the Euler pairing. I see it mentioned in your other question, with different notation. mathoverflow.net/questions/417368/… $\endgroup$
    – Dan Ramras
    Commented Mar 7, 2022 at 15:40
  • $\begingroup$ @DanRamras Yes, thanks! I think on $K^0_{top}(X)$, and vector bundles $E,F$, we can define $\langle E,F\rangle=\chi(E^\vee\otimes F)$, and the definition can be extended to complexes on vector bundles hence $K_{top}^0(X)$, and I am not quite sure if the pairing naturally extend to $K^1_{top}(X)$.. (The motivation was to understand the topological $K$-theory in the case of curves, I had (carelessly )thought such thing exist as an analogue of Poincare duality) $\endgroup$
    – user39380
    Commented Mar 8, 2022 at 20:23
  • $\begingroup$ Just an FYI: For general complex projective manifolds, Section 5.1 in The integral Hodge conjecture for two-dimensional Calabi-Yau categories might be helpful (which refers to the fancy paper Topological K-theory of complex noncommutative spaces). $\endgroup$
    – user479269
    Commented Mar 28, 2022 at 15:08

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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.

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    $\begingroup$ What map $H\mathbb{Z}→KU$? I do not think there's such a map of spectra, let alone of ring spectra (if there were, $KU$ would have trivial k-invariants, which they are not). $\endgroup$
    – Denis Nardin
    Commented Mar 7, 2022 at 15:24
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    $\begingroup$ One might, however, use the map $KU \to \prod_{i\in \mathbb{Z}} \Sigma^{2i} H\mathbb{Q}$ of ring spectra. $\endgroup$
    – Lennart Meier
    Commented Mar 7, 2022 at 15:28
  • $\begingroup$ Hmmm... I guess I was thinking of the map $ku\to H\mathbb{Z}$... I was a little worried that this seemed too simple... Anyway, I think from the OP's other recent post, Euler pairing means something else, so I'll remove this part of the answer. $\endgroup$
    – Dan Ramras
    Commented Mar 7, 2022 at 15:29
  • $\begingroup$ The pairing the OP is talking about is described here mathoverflow.net/questions/417368/… $\endgroup$
    – Dan Ramras
    Commented Mar 7, 2022 at 15:35
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    $\begingroup$ Your formula for $\Sigma M^g$ is not correct (it should be $\Sigma M^g\simeq (\bigvee_{2g}S^2)\vee S^3$). For an explicit homotopy equivalence choose a basis for $H^1M^g$, represented as maps $M^g\rightarrow S^1\simeq K(\mathbb{Z},1)$. Suspend and use the suspension coordinate to add the maps together with the suspension of the pinching map $M^g\rightarrow S^2$. The result is a homology equivalence $\Sigma M^g\simeq\bigvee_{2g} S^2\vee S^3$. (you should also have $[M^g,U]\cong\bigoplus_{2g}\pi_1U$ (in accordance with your first paragraph). $\endgroup$
    – Tyrone
    Commented Mar 7, 2022 at 16:19

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