Let us consider $\mathbb{R}^2$ with coordinates $\{x_1, x_2\},$ endowed with the standard flat Riemannian metric $g.$

There is a standard way to extend a metric to tensors. Namely, given pure tensors $a_1 \otimes \dots \otimes a_n$ and $b_1 \otimes \dots b_n,$ we set $g( a_1 \otimes \dots \otimes a_n, b_1 \otimes \dots b_n):= g(a_1, b_1) g(a_2, b_n) \dots g(a_n, b_n)$ and extend this map to $V^{\otimes n}$ by linearity.

In particular, this means that given $dx_1 \wedge dx_2= dx_1 \otimes dx_2 - dx_2\otimes dx_1,$ then $g( dx_1 \wedge dx_2, dx_1 \wedge dx_2)= 2.$

On the other hand, it is also customary and natural to set $g(dx_1 \wedge dx_2, dx_1 \wedge dx_2 ):= \det(g(dx_i, dx_j)) =1.$

Do these two conventions really conflict, or am I making a mistake somewhere?

  • $\begingroup$ The wedge product is maybe half the tensor-commutator. $\endgroup$ Aug 13, 2017 at 19:38
  • $\begingroup$ @MarianoSuárez-Álvarez : The issue with such a convention is that $dx_1 \wedge dx_2(\partial_{x_1}, \partial_{x_2})$ ought to equal $1$ (an orthonormal basis should evaluate to 1 on a volume form). $\endgroup$
    – user142700
    Aug 13, 2017 at 19:59

1 Answer 1


Yes, they are different inner products. In fact, because there are two different conventions for how the wedge product relates to the tensor product, there are actually three different possible inner products on differential forms.

1. The Pointwise Hodge Inner Product: This is the inner product on differential forms that satisfies $$ \langle\varepsilon^{i_1}\wedge\dots\wedge\varepsilon^{i_k},\varepsilon^{i_1}\wedge\dots\wedge\varepsilon^{i_k}\rangle = 1 $$ whenever $i_1<\dots<i_k$ and $(\varepsilon^1,\dots,\varepsilon^n)$ is a local orthonormal frame for $T^*M$. It's also characterized by $$ \omega \wedge *\eta = \langle\omega,\eta\rangle dV_g, $$ where $*$ is the Hodge star operator and $dV_g$ is the Riemannian volume form. In particular, on $\mathbb R^2$ with the Euclidean metric, it satisfies $$ \langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = 1. $$ This inner product is universally used in Hodge theory.

(The terminology "pointwise Hodge inner product" is not standard, but it's the most accurate name I can think of for this inner product, given that the term Hodge inner product usually refers to the corresponding global inner product on differential forms obtained by integrating this pointwise inner product.)

2. The Tensor Inner Product: This is obtained from the usual inner product on tensors, by viewing the bundle of differential $k$-forms as a subbundle of the bundle of covariant $k$-tensors. The relationship between this and the pointwise Hodge inner product depends on which convention is used for wedge products. In my books, I refer to these conventions as the determinant convention and the Alt convention.

2a. Using the determinant convention: This is the wedge product convention that satisfies $$ \omega^1\wedge\dots \wedge \omega^k(v_1,\dots,v_k) = \det (\omega^j(v_k)) $$ when $\omega^1,\dots,\omega^k$ are covectors and $v_1,\dots,v_k$ are vectors. In particular, with this convention, we have $dx^1\wedge dx^2 = dx^1\otimes dx^2 - dx^2 \otimes dx^1$, so $$ \langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = 2. $$

2b. Using the Alt convention: With this convention, the wedge product is determined by $$ \omega\wedge\eta = \operatorname{Alt} (\omega\otimes \eta) $$ when $\omega$ and $\eta$ are any differential forms whatever. In particular, with this convention, we have $dx^1\wedge dx^2 = \tfrac 1 2 (dx^1\otimes dx^2 - dx^2 \otimes dx^1)$, so $$ \langle dx^1\wedge dx^2,dx^1\wedge dx^2\rangle = \tfrac 1 2. $$

More generally, the tensor inner product is $k!$ times the pointwise Hodge inner product on $k$-forms if you use the determinant convention, and $(1/k!)$ times if you use the Alt convension.

  • $\begingroup$ Which of these would work if we replaced $\mathbb R$ with a field of nonzero characteristic? Clearly 2a would and 2b wouldn't. I can't quite see how to define 1 in that case. $\endgroup$ Aug 14, 2017 at 9:13
  • $\begingroup$ @OscarCunningham -- I have no idea. Does an inner product even make sense in a vector space over a field of characteristic 2? $\endgroup$
    – Jack Lee
    Aug 14, 2017 at 13:45
  • $\begingroup$ I think officially "inner product" is defined only over $\mathbb R$ and $\mathbb C$. I should have said something like "non-degenerate symmetric bilinear form". Really I'm just wondering if we could get any insight into which of these inner products is "most natural" by seeing which ones generalise. In the same vein I've seen people argue that it's more natural to consider $\Lambda^2 V^*$ as a quotient of $V^*\otimes V^*$ rather than a subspace, because in characteristic $2$ the alternating bilinear forms really can't be thought of as a subspace of $V^*\otimes V^*$. $\endgroup$ Aug 14, 2017 at 14:08
  • $\begingroup$ This has been driving me crazy, so it's nice to see that it is confirmed to be different conventions. Have you seen the choice to embed the forms as tensors by the convention: $a\wedge b = \frac1{\sqrt 2} (a\otimes b - b\otimes a)$? $\endgroup$
    – Keshav
    Dec 2, 2021 at 18:33

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