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In this topic: Scalar product on manifold. Henry said $(\omega, \eta)=\int_M \omega \wedge \star \eta=\int_M \left<\omega, \eta \right>\mathrm{d vol}$ is an inner product on forms, where $\left<\cdot, \cdot \right>$ is the pointwise inner product induced by riemannian metric i.e. $\left<\alpha, \beta \right>:=g^{ij}v_iw_j$ for $1$-forms $\alpha=v_i e^i$ and $\beta=w_i e^i$, $\left<\alpha_1 \wedge \ldots \wedge \alpha _p, \beta_1 \wedge \ldots \wedge \beta_p \right>:=\det(\left<\alpha_i, \beta_j \right>)$ etc. I assume $(\omega, \eta):=\int_M \left< \omega, \eta\right> \mathrm{dvol}$ is not an inner product on pseudoriemannian manifold, since $\left<\alpha, \beta \right>:=g^{ij}v_iw_j$ is no longer an inner product. Can we save this structure somehow? Does exist an inner product on differential forms on pseudoriemannian manifolds? ;>

P Lrc
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    Certainly such an inner product always exists. (Just choose a Riemannian metric and use the Hodge inner product with respect to that.) But there's no way to assign such an inner product to every pseudo-Riemannian manifold that will be invariant under isometries. One way to see this is that such an inner product on $1$-forms would determine a pointwise Riemannian metric (for example, by applying the global inner product to forms supported in smaller and smaller neighborhoods and taking a limit). But there's no way to canonically associate a Riemmanian metric to each pseudo-Riemannian one. – Jack Lee Dec 24 '17 at 20:51
  • Hehe, its obvious there's an inner product if I choose a Riemannian metric on manifold. I wasn't enough accurate. I was thinking about pseudoriemannian manifold. Can we define an inner product on pseudoriemannian manifold using its metric? – P Lrc Dec 24 '17 at 23:13
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    My comment was meant to address that. Finding one that depends only on the metric is essentially asking that it be given by some functorial construction that's preserved by isometries. There is no such thing. If you're willing to allow it to depend on more than just the metric, then there are things that can be done. For example, on a Lorentz manifold, if you choose a one-dimensional timelike line bundle, then you can use the Lorentz metric to define a Riemannian metric: it's the negative of the Lorentz metric on the line bundle, and equal to the Lorentz metric on its orthogonal complement. – Jack Lee Dec 24 '17 at 23:41
  • Ok, I understand, but $\left| \omega \right|:=\sqrt{\left|(\omega, \omega)\right|}$ where $(\cdot, \cdot)$ is defined above will be still a norm on pseudo Riemannian manifold, right? – P Lrc Dec 25 '17 at 21:31
  • No, it's only a seminorm, because $|\omega|$ can be zero even if $\omega$ is not identically zero. – Jack Lee Dec 26 '17 at 15:11
  • I've realized it's not a norm (forgive a stupid question), but are you sure it is a seminorm? In particular does it satisfy the triangle inequality? For $$g:=e^1 \otimes e^1 - e^2 \otimes e^2$$ I define $$\left<\alpha, \beta \right>=\left<v_i e^i, w_j e^j \right>:=g^{ij}v_iw_j$$ and $$\left|\alpha \right|:= \sqrt{\left|\left< \alpha, \alpha \right>\right|}= \sqrt{\left|\sum_i g^{ii}v_i^2 \right|}=\sqrt{\left|v_1^2-v_2^2 \right|}$$ Alas, the triangle inequality is not satisfied for $\alpha:=ae^1+ae^2$ and $\beta:=ae^1-ae^2$, where $a>0$. :( – P Lrc Dec 26 '17 at 20:06
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    Oh, you're right. It's not even a seminorm. Sorry for the misleading comment. – Jack Lee Dec 26 '17 at 20:08

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