# Inner product on diffential forms independence ortonormal basis

Suppose $$\{e_1,...,e_n\}$$ is a positive orthonormal basis for the tangent space at a point $$p$$ in an oriented n-manifold $$M$$, then define the inner product on $$\Omega^k(M)$$, for each $$k$$, by: $$\langle\,\omega,\tau\rangle_p=\sum_{i_1<... The claim is that this inner product independs on the choice of orthonormal basis for $$T_pM$$ the tangent space. It's clear to me it suffices to show this holds for basis elements of each space. Also, it's easy in the case $$k = n$$, because if we take another positive orthonormal basis then the form applied at this new basis will be equal to the form applied at the first basis times the determinant of the change of basis matrix, which will be $$1$$. However, for $$k\neq n$$, some subdeterminants will appear, then I'm not sure what to do...

Scalar product on vector space can be viewed as a map $$\langle \cdot, \cdot\rangle : V \to V^*$$ such that $$x\mapsto \langle x, \cdot\rangle$$, the scalar product you have is just $$k$$-th exterior power of this map. $$\langle \cdot, \cdot \rangle_p=\bigwedge^k \langle \cdot, \cdot \rangle_p$$ and it acts just like you've said.

This algebraical definition is given indepentent of basis and coincides with yours. If you are familiar with categories and functors you can notice, that exterior power is a functor.

• I'm aware of that, however I wanted an explicit proof that different basis induce the same inner product... May 15 '19 at 15:13
• The definitions coincide and my doesn't use any prechosen basis, so other basis cannot define other inner product. May 15 '19 at 15:16
• I guess I'm not that aware of the definitions you just used then. What would be the algebraical definition of this inner product? Wouldn't it depend on the coice of inner product? May 15 '19 at 15:53
• Yes, of course. And your definition also depends, because you use an orthonormal basis. May 15 '19 at 16:01
• Ahhh, I see, it makes sense May 15 '19 at 16:28