2
$\begingroup$

Let $(X,g)$ be a compact smooth Riemannian manifold. It is known that $H^1(X, \mathbb R)\cong \mathrm{Hom} (\pi_1(X), \mathbb R)$, namely there is a natural pairing $$ H^1(X) \times \pi_1(X) \to \mathbb R $$ which is denoted by $\langle,\rangle$. Now, let $\alpha\in H^1(X)$ and $[\gamma]\in\pi_1(X)$. Take a representative loop $\gamma$. And, the metric $g$ naturally induces a norm $\|\cdot\|_g$ on $H^1(X)$ by taking the infimum of all $\|a\|_{L^\infty;g}$ over all 1-forms $a$ representing $\alpha$. We use $L_g$ to denote the length with respect to $g$. Then my guess is as follows:

Question. Can we find a constant $c$ so that $$ \langle \alpha, [\gamma] \rangle \le c \|\alpha\|_g \cdot L_g(\gamma) $$ for any $\alpha$ and $[\gamma]$? If so, can we simply take $c=1$?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Yes, directly from the definitions, is not it? $\endgroup$
    – Anton Petrunin
    Commented Dec 14, 2018 at 5:17
  • $\begingroup$ @AntonPetrunin Thank you for comment. I believe this is sort of trivial but I still miss some point. I just naively get the pairing from the universal coefficient theory which is purely topological, and I don't see the way to relate to a metric. $\endgroup$
    – Hang
    Commented Dec 14, 2018 at 5:24
  • 6
    $\begingroup$ $\langle\alpha,[\gamma]\rangle=\int_\gamma\alpha=\int\alpha(\gamma')\le \int |\gamma'(t)|\cdot |\alpha_{\gamma(t)}|\cdot dt\le \|\alpha\|_g\cdot \mathop{\rm length}_g\gamma$. $\endgroup$
    – Anton Petrunin
    Commented Dec 14, 2018 at 5:54

0

Reset to default