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The signature of compact oriented $4k$-manifolds has the following additivity property, first observed by S. P. Novikov: If two manifolds are glued by an orientation-preserving diffeomorphism of their boundaries, then the signature of their union is the sum of their signatures. C.T.C. Wall showed that this additivity property does not hold for the more general situation where one glues two manifolds along a common submanifold of the boundaries, which itself has a boundary.

My questions is: does it hold when the manifolds has several boundary components and we only glue part of them? (which means the final manifold still has boundary)

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2 Answers 2

up vote 8 down vote accepted

Yes, the signature is additive in this case. Wall gave a precise description of the failure of additivity in gluing two $4k$-manifolds along a subset of their boundaries. (I think the name of the paper is "On the non-additivity of the signature", or something like that, but I don't have it in front of me.) Let $A$ and $B$ be two $4k$-manifolds, and suppose we are gluing along $S\subset \partial A$ and an identical $(4k{-}1)$-manifold $S\subset \partial B$. Consider the $(4k{-}2)$-manifold $\partial S$. It bounds three manifolds: $S$, $\partial A\setminus S$, and $\partial B\setminus S$. The kernels of these inclusions give rise to three lagrangian subspaces of the symplectic middle-dimensional homology of $\partial S$; call them $L_1, L_2, L_3$. Then $\sigma(A \cup_S B) = \sigma(A) + \sigma(B) + c(L_1, L_2, L_3)$, where $\sigma$ denotes signature and $c(L_1, L_2, L_3)$ is an integer which, as the notation suggests, depends only on the three lagrangians $L_1$, $L_2$, and $L_3$.

You assume that $\partial S$ is empty. In this case $c(L_1, L_2, L_3) = 0$ and the signature is additive.


One simple way to define $c(L_1, L_2, L_3)$ (not the way Wall does it) is as follows. Let $n$ be the dimension of $L_i$ and let $V$ be the symplectic vector space which contains the lagrangians. If $n=1$ we define $c(L_1, L_2, L_3) = \pm 1$ according to the cyclic order of $L_1$, $L_2$, and $L_3$, assuming the three lagrangians are distinct. If they are not distinct we define $c(L_1, L_2, L_3) =0$. For $n>1$ it is not hard to show that we can find a symplecitc direct sum decomposition $V = \oplus_\alpha V^\alpha$, with $V^\alpha$ 2-real-dimensional, and lagrangians $L_i^\alpha\subset V^\alpha$, such that $L_i = \oplus_\alpha L_i^\alpha$. We now extend linearly and define $c(L_1, L_2, L_3) = \sum_\alpha c(L_1^\alpha, L_2^\alpha, L_3^\alpha)$.

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The answer is yes, and it is an immediate consequence of the Atiyah-Patodi-Singer index theorem, more precisely Theorem 4.14. of

M.F. Atiyah, V.K. Patodi, I.M. Singer, Spectral asymmetry and Riemannian geometry I, Math. Proc. Camb. Phil. Soc. vol. 77, 1975, 43-69.

Admittedly the above argument feels like mouse-hunting with a bazooka. There are more elementary ways to see this but they would involve a bit more sweat.

Update. For an elementary proof of the this general form of additivity see Theorem 27.5 in

B.A. Dubrovin, A.T. Fomenko, S.P. Novikov, Modern Geometry-Methods and Applications. Part III. Introduction to Homology Theory, Graduate Texts in Math., vol. 124, Springer Verlag, 1990.

The above Novikov is the same Novikov in Novikov additivity, and the proof in the above reference is Novikov's original argument which was also sketched in Kevin Walker's answer.

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Thanks Liviu, I can't find Novikov's origional proof. I heard that Atiyah-Singer gave a proof in their 1968 paper: The index of elliptic operators. III. I can't access that paper now. Do you know whether it also implies the claim trivially? So we don't have to use the bazooka. –  J. GE Sep 3 '12 at 13:33
    
See this related question : mathoverflow.net/questions/69167/… –  BS. Sep 3 '12 at 13:43
    
@ GB See the update. –  Liviu Nicolaescu Sep 3 '12 at 14:42
    
+1 for hunting mice with bazookas. –  Vidit Nanda Sep 3 '12 at 21:38

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