-1
$\begingroup$

Let $M$ be a connected closed manifold of dimension $n\geq 2$.

Can it happen that for any $I\subsetneq I_n=\{1, \dots, n\}$ there are continuous maps $f, g:M\to M$ such that $f^*=g^*|_{\oplus_{i\in I}H^i(M, \mathbb{Z})}$ yet $f^*\neq g^*|_{\oplus_{i\in I_n}H^i(M, \mathbb{Z})}$?

Improve this question
$\endgroup$
4
  • $\begingroup$ If it is true for each one-element subset, then it is true for the whole set, since the action on the direct sum is the direct sum of actions, no? $\endgroup$
    – Vladimir Dotsenko
    Commented Mar 30, 2021 at 10:18
  • $\begingroup$ @VladimirDotsenko I'm not sure I understand you correctly. A priori there is $2^n-1$ subsets you should check this on but it is enough to check on $n$ subsets (each omitting one index). Is that what you meant? $\endgroup$
    – klaus
    Commented Mar 30, 2021 at 10:40
  • 1
    $\begingroup$ I think the misconception is that the OP wanted to allow that $f,g$ depend on the choice of $I$, but V.D. assumed the same functions should provide an example for all choices? $\endgroup$
    – Jens Reinhold
    Commented Mar 30, 2021 at 11:00
  • $\begingroup$ @klaus I see what you meant. For some reason, the wording you use (while being correct) created the wrong quantifiers in my head, as Jens Reinhold remarks. $\endgroup$
    – Vladimir Dotsenko
    Commented Mar 30, 2021 at 12:51

1 Answer 1

Reset to default
1
$\begingroup$

I believe this cannot happen: trivially, if $\text{H}^1(M;\mathbb Z) = 0$, you clearly won't find a example of $f, g$ as desired with $I = \{2, \dots, n\}$.

Suppose first that $M$ is orientable. Pick $0 \neq \alpha \in \text{H}^1(M)$ that is not a proper integral multiple of some other cohomology class. Then, by Poincaré duality, we can find $\beta \in \text{H}^{n-1}(M;\mathbb Z)$ such that $\alpha \beta = \mu_M$, the dual of the fundamental class $[M]$. Hence if $f, g \colon M \to M$ are such that $f_{\ast} = g_{\ast}$ in degrees below $n$, we also get $f_{\ast} \mu_M = (f_{\ast} \alpha)(f_{\ast} \beta) = (g_{\ast} \alpha)(g_{\ast} \beta) = g_{\ast} \mu_M$ and thus $f_{\ast} = g_{\ast}$ in all degrees; so for $I = \{1, \dots, n-1\}$ no $f,g$ as desired exist.

If $M$ is non-orientable, the same argument with $\mathbb F_2$ coefficients should work.

Improve this answer
$\endgroup$
2
  • $\begingroup$ if $M$ is non-orientable don't you get $H^n=\mathbb{Z}/2$? $\endgroup$
    – klaus
    Commented Mar 30, 2021 at 11:04
  • $\begingroup$ Of course, thanks. I edited the answer, hope it works now. $\endgroup$
    – Jens Reinhold
    Commented Mar 30, 2021 at 11:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .