MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am studying SO(3) bundles on K3 surfaces, and in particular I want to understand several remarks made by Vafa and Witten in their article on S-duality in supersymmetric Yang-Mills theories. As is well known, SO(3) bundles are fully determined by their second Stiefel-Whitney class $w_2(E) \in H^2(K3,\mathbb{Z}_2)$ and the instanton number, which in this case can be written as $\frac14p_1$.

Now Vafa-Witten remark on page 55 of their article that K3 has a very large diffeomorphism group, which permute the possible values of $w_2$; they say that 'an obvious' diffeomorphism invariant of $w_2$ is $w_2^2\mod4$. But:

Why does $w_2$ change at all under a diffeomorphism? Any diffeomorphism $f$ from some space $X$ to K3 induces the pullback bundle $f^*E$ over $X$, but by the naturality of $w_2$, it would remain the same.

Then later on they say that modulo 2 the intersection form on $H^2(K3,\mathbb{Z})$ is equivalent to 11 copies of $H = \begin{pmatrix}0&1\\\\1&0\end{pmatrix}$. Any idea why this is the case? (The intersection form of K3 is of the form $2(-E_8)\oplus 3H$.)

share|cite|improve this question
Hi Manuel. Naturality says that $f^*w_2(E)=w_2(f^*E)$, but it is quite possible that this doesn't equal $w_2(E)$. Also, I think they are talking about self-diffeomorphisms of $K3$ (which may permute summands of $H^2(K3)$). – Mark Grant Mar 26 '11 at 12:31
Concerning $E_8$, modulo 2 we have $2(-E_8) = E_8 \oplus (-E_8)$ which is in turn isomorphic to $8H$. – Bruno Martelli Mar 26 '11 at 16:06
Instead of $SO(3)$ bundles on $K3$, consider $O(1)$ bundles on $T^2 = \{ (x,y) \}$, classified by $w_1$. Consider the bundle that flips orientation around the $x$ circle and not the $y$, so $w_1 = (1,0) \in H^1(T^2; Z_2)$. If you consider the diffeomorphism $(x,y)\mapsto (y,x)$ of $T^2$, the new $w_1 = (0,1)$. – Allen Knutson Mar 26 '11 at 21:15
Well, the intersection form is even, so you should precise what modulo 2 means, presumably $H^2/2H^2$ (as implicit in Bruno Martelli's remark). – BS. Apr 1 '11 at 9:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.