# Are the following two inner products on differential forms equal?

There are two inner product on differential forms:

1. $$\langle \alpha,\beta\rangle$$ induced from Riemannian metric $$g$$ by defining on 1-forms as dual of vector fields then extending to all differential forms i.e. $$\langle e_{i_1}\wedge \dots\wedge e_{i_k}, e_{j_1}\wedge \dots\wedge e_{j_k}\rangle =\det[\langle e_{i_s}, e_{j_s}\rangle]$$.
2. On compact oriented Riemannian manifold $$(M,g)$$ $$(\alpha,\beta)=\int_M\alpha\wedge\star\beta.$$

Q1: Are these two inner products on differential forms equal?

Q2: If the answer to Q1 is "NO" then is it important to notice that two operators are adjoint of each other (or an operator is symmetric or self-adjoint) w.r.t. which metric? e.g. $$d$$ and $$\delta$$ that are adjoint w.r.t. second inner product but I don't know it is w.r.t. other one.

• How do you extend the first one to differential form? Mar 14 at 17:31
• No, the first definition is not on differential forms; it's pointwise or at best local. I suggest, however, that you relate $\alpha\wedge\star\beta$ at a point to the definition in 1. Mar 14 at 21:31
• I think what @TedShifrin meant was that (1) is defined as an inner product on $\Lambda^p_xM$ for $x\in M$, while (2) is defined as an inner product on $\Gamma(\Lambda^pM)$. The relation between (1) and (2) is that (2) is the global version of (1), obtained by integrating (1). Mar 15 at 9:34
• $\Gamma(\Lambda^pM)$ is the set of sections of the vector bundle $\Lambda^pM$, and is usually written as $\Omega^p(M)$ yes. Its elements are differential forms of degree $p$, and are global objects. Mar 15 at 10:31
• What I said is that $(2)$ is not a Riemannian metric. It is an inner product on the vector space of $p$-forms. An analog would be the inner product on $L^2[0,1]$ defined by $\langle f,g\rangle = \int_0^1 fg$. There is no such thing as $\left(\int_0^1f^2\right)(x)$, because $\int_0^1f^2$ is a number, not a function. Mar 19 at 11:29

Disclaimer: the comment section is overgrowing so here is an answer that I hope will erase all your doubts.

First part: As said in the comment section, $$(1)$$ is a inner product on the vector space $$\Lambda^p_xM$$ while $$(2)$$ is an inner product on the vector space $$\Omega^p(M)$$. The link between them is that $$(2)$$ is obtained by integrating $$(1)$$ over the whole manifold $$M$$.

An analog is this: consider the set $$M=[0,1]$$. Then for each $$x\in [0,1], T_x[0,1] = \mathbb{R}$$ and one can define an inner product on $$T_x[0,1] = \mathbb{R}$$ by $$\langle a,b \rangle_x = a\times b$$. This is $$(1)$$.

A vector field on $$[0,1]$$ is just a smooth function $$f:[0,1] \to \mathbb{R}$$, and $$(2)$$ is here $$\langle f,g\rangle = \int_0^1 f(x) g(x) \mathrm{d}x = \int_0^1 \langle f, g\rangle_x \mathrm{d}x.$$

Second part: If $$V$$ is a vector space and $$\langle\cdot,\cdot\rangle$$ is an inner product, one can create a Riemannian metric on $$V$$, thought as a manifold the following way. As a vector space, the tangent bundle of $$V$$ is trivial: $$TV = V\times V$$ and one can define the Riemannian metric $$g_v = \langle\cdot,\cdot\rangle$$ for $$v\in V$$. It is a constant Riemannian metric because the canonical trivialization makes $$g_v$$ to be a function independant of $$v \in V$$. Take $$V = \Omega^p(M)$$ and $$\langle \alpha,\beta\rangle = \int_M \alpha \wedge \star \beta$$. Now, forget that $$V$$ and $$\langle\cdot,\cdot\rangle$$ is defined thanks to a Riemannian manifold $$(M,g)$$ and just look at its structure: it is a vector space with an inner product. Hence, for this inner product $$\|\alpha\|$$ is a number.

If you really want to think of this construction as a Riemannian manifold, like in the first paragraph, then $$\|\alpha\|$$ will be a function: $$\|\alpha\| : \beta \in \Omega(M)^p \mapsto \|\alpha\|(\beta) = \|\alpha\|\in \mathbb{R}$$ which is constant and does not take points of $$M$$ as entries.

Comment: if you really do not understand what I said, here is just a question for you: for $$x \in M$$, how would you define $$\left(\int_M \alpha\wedge \star \beta\right)(x)$$?

This is the exact same thing as this question: how would you define $$\left(\int_0^1 t^2 \mathrm{d}t\right)\left(\frac{1}{2}\right)$$?