[Math] Symmetric and wedge product in algebra and differential geometry

Question

Which is the correct identity?

1. $dx \, dy = dx \otimes dy + dy \otimes dx$ $~~~$or$~~~$ $dx \, dy = \dfrac{dx \otimes dy + dy \otimes dx}{2}~$?
2. $dx \wedge dy=dx \otimes dy – dy \otimes dx$ $~~~$or$~~~$ $dx \wedge dy=\dfrac{dx \otimes dy – dy \otimes dx}{2}~$?

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Here is my understanding of the question from the point of view of:

Linear algebra:

Let $V$ be a vector space. The symmetric algebra $S(V)$ is a quotient of the tensor algebra $T(V)$. The symmetric product $v \cdot w$ of elements of $V$ does not make sense a priori in $T(V)$, but one can identify $S(V)$ with the space of symmetric tensors, which is a subspace of $T(V)$ where the restriction of the projection map $T(V) \to S(V)$ is an isomorphism. Under this isomorphism, the symmetric product $v \cdot w$ corresponds to the element $\dfrac{v \otimes w + w \otimes v}{2}$ of $T(V)$. Same story for the exterior algebra $\Lambda(V)$ and alternating tensors: the wedge product $v \wedge w$ is identified with the alternating tensor $\dfrac{v \otimes w – w \otimes v}{2}$.

So contrary to what I have read in some places (e.g. accepted answer here), in my opinion there is one natural way to identify symmetric products to symmetric tensors (resp. wedge products to alternating tensors)¹. Conclusion: at least from the algebraic point of view, it seems to me that the natural thing to say is:

1. $dx \, dy = \dfrac{dx \otimes dy + dy \otimes dx}{2}$
2. $dx \wedge dy = \dfrac{dx \otimes dy – dy \otimes dx}{2}$

Differential geometry:

Again, I feel like there is only one choice we want to make here, contrary to what I have read sometimes:

1. $dx \, dy = \dfrac{dx \otimes dy + dy \otimes dx}{2}$, because $dxdx + dydy = dx^2 + dy^2 $ should be the standard metric (or inner product) on $\mathbb{R}^2$ (who would want $dx^2 + dy^2$ to mean something else?)
2. $dx \wedge dy = dx \otimes dy – dy \otimes dx$ because $dx \wedge dy$ should be the standard area form (or determinant) on $\mathbb{R}^2$ (again, who would want $dx \wedge dy$ to mean something else² ?).

Unfortunately, the answer 2. is different than what we found from the algebraic point of view. Worse, the choices made for the symmetric product and the wedge product do not seem to be consistent.

Does anyone feel like they have a satisfying way to understand this issue?

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¹ as I have tried to explain briefly. Said differently, it is natural to ask that the identification $\mathrm{Sym}^2 V \stackrel{\sim}{\to} S^2V$ should be the restriction of the projection map $p: V\otimes V \to S^2V$. (Same story for the wedge product).

² Said differently, when one defines integration of differential forms, integrating $f(x, y)\, dx \wedge dy$ should produce the Lebesgue integral $\int f(x,y) dx\,dy$. I don't think anyone uses a different convention (?). Other remark: in complex differential geometry, I find the most natural identity between a Kähler Hermitian metric $h$, the Riemannian metric $g$ and the Kähler form $\omega$ to be $h = g – i\omega$. Try $h = dz \otimes d\overline{z}$: then $g = dx \otimes dx + dy \otimes dy$ and $\omega = dx \otimes dy – dy \otimes dx$. It is nice to write $g = dx^2 + dy^2$ and $\omega = dx \wedge dy$, in particular, the Kähler form is the area form of the Riemannian metric.

Answer 1

The motivation for the coefficient $\frac{1}{n!}$ is as follows : if $f : V\times...\times V\to\mathbb{K}$ is n-linear and alternate form, we define the alternator $\mathrm{Alt}$ so that $\mathrm{Alt}(f)=f$. That is $$f(x_1,...,x_n)=\frac{1}{n!}\sum_{\sigma\in S_n}\varepsilon(\sigma)f(x_{\sigma(1)},...x_{\sigma(n)})=\mathrm{Alt}(f)(x_1,...,x_n).$$