Subject: Response to Jim Stasheff's question
Date: Tue, 30 Apr 2002 13:54:25 -0400 (EDT)
From: Adam Sikora <sikora@CRM.UMontreal.CA>
To: Don Davis <dmd1@lehigh.edu>

Dear Prof. Davis,

I am enclosing a response to Jim Stasheff's question:

Thm. If G=Z/2 acts on a closed orientable mfld X by reversing its
orientation and dim X=n, Y=X/G then b_k(X)=b_k(Y)+b_{n-k}(Y).

Note that the G-action does not need to be free.

Proof: Denote the generator of G by g.
Let H_i^k, H_a^k for k=0,1,2,... denote the invariant and
anti-invariant part of H^k(X;Q) with respect to G-action.
(g fixes H_i^k, and gx=-x for x in H_a^k).
H^k(X)=H_i^k+H_a^k and the cup product on H^*(X) restricts to
nondegenerate cup product
H_i^{n-k} x H_a^k -> H^n(X).
(Proof: If x is in H_i^{n-k} then there exist y_i, y_a in H_i^k and H_a^k
respectively such that x u (y_i+y_a)=w \ne 0 in H^n(X). After applying g,
we get x u (y_i-y_a)=-w. Hence x u y_a=w).
Hence dim H_a^k=dim H_i^{n-k}.
Now the statement follows the fact that H^k(Y)=H_i^k, see
eg. Seminar on Transformation Groups by A. Borel, Princ. Univ. Press 1960.
Corollary III.2.3 page 38.

Best Regards,

Adam Sikora