Talk:Novikov conjecture

What does ${\mathcal {L}}_{i}(M)$ stand for? - 69.174.134.88 21:50, 9 June 2006 (UTC)

Glaring error

We supposedly have:

$f:M\rightarrow BG$

$x\in H^{n-4i}(M;\mathbb {Q} ).$

$\left\langle f^{*}(x)\cup L_{i}(M),[M]\right\rangle \in \mathbb {Q}$

How can we take $f^{*}(x)$? Cohomology is contravariant, so $f^{*}:H^{*}(BG)\rightarrow H^{*}(M)$. This mistake renders the statement of the conjecture totally non-sensical.

In the linked seminar notes there ist the same mistake. I think, x should be in H^{n-4i}(BG). This makes most sense in view of how a result of Novikov is shown to be a special case of the Novikov conjecture the seminar notes, I think. --Cardano 13:09, 12 October 2006 (UTC)

That link (as well as the mistake) was on the original page, created a year ago. Thank you, anonymous person. And Cardano you're right - the conjecture is stated in the referreed article, page 195 at http://www.maths.ed.ac.uk/~aar/books/wall1.pdf Orthografer 15:16, 12 October 2006 (UTC)