Apparently I cannot get the right sign in deriving classical Gauss-Bonnet from generalised one!

According to many references (e.g. Madsen and Tornehave, Nakahara, Milnor and Stasheff if you don't use their particular convention about orientation) generalised Gauss-Bonnet is

$\int_M Pf(-R/2\pi)=\chi(M)$

where $R$ is the curvature 2-form and the Pfaffian of a 2lx2l antisymmetric matrix is $Pf(A)=\frac{1}{2^ll!}\epsilon ^{i_1 \ldots i_{2l}}A_{i_1 i_2} \ldots A_{i_{2l-1}i_{2l}} \ $ i.e. the sign convention is such that for the matrix A={{0,a},{-a,0}} Pf(A)=a.

For l=1, M is a surface and $R^a_{\phantom{a}b}=\frac{1}{2}R^a_{\phantom{a}bcd}e^c\wedge e^d =K (g_{ac}g_{bd}-g_{ad}g_{bc})$ where $K$ is the Gaussian curvature, $g_{ab}$ the metric tensor and $\{e^a\}$ an orthonormal basis. For the Pfaffian I get $Pf(-R/2 \pi)= -\frac{1}{4\pi} 2 \epsilon_{12} R^1_2= -\frac{1}{2\pi} R^1_{\phantom{1}2}=-\frac{1}{2\pi}R^1_{\phantom{1}212}e^1\wedge e^2$

$e^1\wedge e^2 $ is the volume form and $R^1_{\phantom{1}212}=K$ therefore

$\int_M Pf(-R/2\pi)=\int_M-K/2\pi=\chi$

which obviously has the wrong sign! I have carefully checked the definitions in the books and I think I am using the same as they use - so please be very explicit in pointing aout where the mistake is! Thanks

  • 5
    $\begingroup$ A common place to make a sign error is the curvature. Compute the curvature of a 2-sphere using your formulas and make sure it's positive. $\endgroup$ – Deane Yang Mar 22 '13 at 12:52
  • $\begingroup$ It turned out that was the problem indeed: my convention for the definition of curvature 2-form where different from those used in the books, resulting in a minus sign difference. $\endgroup$ – GFR Mar 22 '13 at 15:34

Check Example 2.2.63 page 57 and pages 162-163 of these notes to see the correct sign conventions. The usual definition of the pfaffianof a $2\ell\times 2\ell$ skew-symmetric matrix is $(-1)^\ell \times $ your definition of the pfaffian; see pages 56-57 of the same notes.

  • $\begingroup$ Thanks Liviu, I agree that with the conventions that you pointed out everything works out fine. I still have not accepted the answer to see if anyone can confirm that there is a mistake in the books I have mentioned or - more likely - point out what I missed out. In many sources the definition of the Pfaffian is given as $l!A\wedge \ldots \wedge A=Pf(A) vol$ which is equivalent to my definition above. $\endgroup$ – GFR Mar 22 '13 at 12:40
  • 1
    $\begingroup$ Milnor an Stasheff define the pfaffian as you do but the Gauss-Bonnet formula in their case (p.311 in their book) is obtained by integrating $Pf(R/2\pi)$ and not $Pf(-R/2\pi)$ as in your case. This takes care of the the factor $(-1)^\ell$. $\endgroup$ – Liviu Nicolaescu Mar 22 '13 at 13:44
  • $\begingroup$ That is true but hey also use an unconventional orientation: see page 304, note on signs. Basically if the manifold has dimension 2l and l is even their orientation is the same as the conventional one: $e^1\wedge \ldots \wedge e^{2l}$ but otherwise there is a minus sign. Quoting from the book: "Readers who prefer to use the classical sign conventions as in [Spanier], [Warner] and [Bott-Chern] can forget about these signs, but should replace K with -K whenever it occurs in our characteristic class formulas". $\endgroup$ – GFR Mar 22 '13 at 15:21
  • $\begingroup$ I think I have found where the sign comes from: I was using conventions such that the connection form $\omega$ satisfies $de^i+\omega^i_{\phantom{i}j}\wedge e^j=0$, while Milnor-Stasheff has a minus sign, see pag.302. This basically correspond to replace $R^i_{\phantom{i}j} $ with $\R^j_{\phantom{i}i}=-R^i_{\phantom{i}j}$, so that an overall minus results if there is an odd number of $R^i_{\phantom{i}j}$s. $\endgroup$ – GFR Mar 22 '13 at 15:27
  • $\begingroup$ In other words, Deane's comment was exactly what you needed!! $\endgroup$ – YangMills Mar 22 '13 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.