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$~~~$ $dx \wedge dy=\dfrac{dx \otimes dy - dy \otimes dx}{2}~$?

$$$$

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?

$$$$

---

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 $$f(x_1,...,x_n)=\frac{1}{n!}\sum_{\sigma\in S_n}\varepsilon(\sigma)f(x_{\sigma(1)},...x_{\sigma(n)})=\mathrm{Alt}(f)(x_1,...,x_n).$$

Answer 2

Because the similarity between symmetric tensor product and wedge product, I will discuss only the wedge product here. It is common to see both two definitions of wedge product in different textbooks.

Given two differential forms $\alpha\in\bigwedge^p(V)$ and $\beta\in\bigwedge^q(V)$, we can define the wedge product as

1. $\displaystyle \alpha\wedge\beta=\operatorname{Alt}(\alpha\otimes\beta)$

This definition, which is consistent with your algebraic point of view, apparently has its own algebraic simplicity. So it is usually used to derive properties and identities about differential forms.

2. $\displaystyle \alpha\wedge\beta=\frac{(p+q)!}{p!q!}\operatorname{Alt}(\alpha\otimes\beta)$

This definition is mostly seen in physics literature because it's more nature to correspond to physical meaning, especially for volume. Another advantage of this definition is when computing inner product of two $p$-forms, then it has an elegant expression $$\left< u_1\wedge\cdots\wedge u_p, v_1\wedge\cdots\wedge v_p\right>=\det(\left<u_i,v_j\right>)$$ without an annoying coefficient before the determinant, if the first definition is used.

Essentially, two definitions of wedge product, or say exterior product, give two different exterior algebras, but only one up to algebraic isomorphism. In the point of view of category theory, exterior algebra can be defined using universal property, which means no matter how one defines the exterior product for the algebra, without further imposing more constraints, then all these algebras can be isomorphically mapping to each other.

So, I will not regard different definitions as a bother since they are the same thing and there's no canonical way to give a canonical definition. Just like you don't care whether an angle expresses as degrees or radian.

Answer 3

$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\End{\mathop{\mathrm{End}}\nolimits} \newcommand\K{\mathbb K} \newcommand\Cl{\mathop{\mathrm{Cl}}} \newcommand\tr{\mathop{\mathrm{tr}}} $The exterior algebra $\Ext V$ can be completely characterized by its universal property:

• Let $A$ be a unital associative algebra. Then every linear $f : V \to A$ such that $f(v)^2 = 0$ for all $v\in V$ extends uniquely to a homomorphism $\Ext V \to A$, meaning in particular $$f(v_1\wedge\dotsb\wedge v_k) = f(v_1)\dotsb f(v_k)$$ for all $v_1,\dotsc,v_k\in V$.

Let $\K$ be our field of scalars. (Any field, unless I say otherwise!) The bilinear pairing $\Psi : \Ext V^*\times\Ext V \to \mathbb K$ $$ \Psi(\omega_1\wedge\dotsb\wedge\omega_k,\: v_1\wedge\dotsb\wedge v_l) = \delta_{kl}\det\Bigl(\omega_i(v_j)\Bigr)_{i,j=1}^k. $$ can be derived just from the above universal property. I know of three ways to do so, all of which give the same exact $\Psi$! In this sense it is a structure intrinsic to the exterior algebras.

Once we have this, and we embed $\Ext V$ in the tensor algebra in the way you say is natural, we are forced to embed $\Ext V^*$ the other way without the $1/k!$ coefficients if we want the tensor pairing to give us $\Psi$.

---

1. Every $\omega \in V^*$ extends uniquely to an antiderivation $i_\omega$ on $\Ext V$. So we have a map $V^* \to \End_\K({\bigwedge}V)$ which extends to ${\bigwedge}V^* \to \End_\K({\bigwedge}V)$, giving the interior product $\Omega\mathbin\lrcorner X$ for $\Omega \in \Ext V^*$ and $X \in \Ext V$. The pairing is now the scalar part $\Psi(\Omega, X) = \langle\Omega\mathbin\lrcorner X\rangle_0$. This can be found in Werner Greub's Multilinear Algebra.

2. The universal property applied to the diagonal map $V \to V\oplus V \subseteq \Ext(V\oplus V)$ gives us $$\pi' : \Ext V \to \Ext (V\oplus V) \cong \Ext V\mathbin{\hat\otimes}\Ext V,$$ with the isomorphism being the canonical one from the universal property and $\hat\otimes$ the graded tensor product. Now for each $\Omega,\Phi \in (\Ext V)^*$ we get $$\pi\circ(\Omega\mathbin{\hat\otimes}\Phi)\circ\pi' \in (\Ext V)^*$$ with $\pi$ the multiplication on $\Ext V$, making $(\Ext V)^*$ an algebra. We have a map $V^* \to (\Ext V)^*$ by declaring the action of a covector on a $k$-vector, $k\ne1$, to be zero. The universal property gives us $\Ext V^* \to (\Ext V)^*$, which is $\Psi$. This also reveals that $\Psi$ is naturally an algebra isomorphism. This can be found in Quadratic Mappings and Clifford Algebras by Artibano Micali and Jacques Helmstetter.

3. This requires $\K$ to have characteristic $\ne2$. Let $W = V^*\oplus V$. The series of isomorphisms $$W^* \cong (V^*)^*\oplus V^* \cong V\oplus V^* \cong W$$ gives us a bilinear form $\beta$, so we have a Clifford algebra $\Cl(W)$. We have $V^*$ and $V$ as totally isotropic subspaces of $W$, so the universal properties of the exterior algebras and $w^2 = \beta(w,w)$ for all $w \in W \subseteq \Cl(W)$ give us canonical maps of $\Ext V^*$ and $\Ext V$ into $\Cl(W)$. Now define $\tr(X)$ to be the trace of $Y \mapsto XY$ in $\Cl(W)$. The pairing is $\Psi(\Omega, X) = \tr(\Omega X)/\tr(1)$, or in other words the trace of $Y \mapsto \Omega XY$ divided by $\dim\Cl(W)$. This is my own derivation; I've not seen it elsewhere but I can't say no one else has done it.

The Clifford algebra $\Cl(V; q)$ of a quadratic form $q : V \to \K$ is fully characterized by the following universal property:

• Let $A$ be a unital associative algebra. Then every linear $f : V \to A$ such that $f(v)^2 = q(v)$ for all $v\in V$ extends uniquely to a homomorphism $\Cl(V; q) \to A$, meaning in particular $$f(v_1\dotsb v_k) = f(v_1)\dotsb f(v_k)$$ for all $v_1,\dotsc,v_k\in V$.

In (3) above we are using $q(w) = \beta(w,w)$ for $w\in W$.

Answer 4

I think also algebraically there can be reasons for different conventions. One can either want a product on alternating tensors which corresponds to the wedge product on $\Lambda(V^*)$ or one for which the contractions $\iota_v$ are derivations. These two properties cannot be achieved both at the same time, so different convention arise depending on which property one decides to maintain.

Some more details: Let $V$ be a vector space and $V^*$ its dual. Then the following structures are naturally available:

• The wedge product on the quotient $\Lambda( V^*)$ of $T(V^*)$ induced by the (juxtaposition) product on the tensor algebra $T(V^*)$.
• The linear subspace of alternating tensors $Alt\subset T(V^*)$. The quotient map $T(V^*)\rightarrow \Lambda(V^*)$ restricts in characteristic 0 to an isomorphism $Alt\rightarrow \Lambda(V^*)$.
• A degree -1 contraction $\iota_v$ on $T(V^*)$ for each $v\in V$. These are derivations for the product on $T(V^*)$ and they map the alternating tensors $Alt\subset T(V^*)$ to themselves.

One may now attempt to define a product on $Alt$ such that:

a) The product on $Alt$ corresponds to the wedge product under the projection to $\Lambda(V^*)$. This works only in characteristic 0. (Contractions $\iota_v$ are then derivations only after a degreewise rescaling).

b) The product on $Alt$ is such that the contractions $\iota_v$ are graded derivations of the product. (This product on $Alt$ corresponds then to that on $\Lambda(V^*)$ only after a degreewise rescaling of the projection map).

Each of these two requirements (essentially) determine a unique (graded commutative) product on $Alt$, but they do not determine the same product. A parallel discussion applies to symmetric tensors $Sym$ and the symmetric algebra $S(V^*)$.

In the conventions preferred in differential geometry (by you and many others), one gives up, for alternating forms, the compatibility of the product with the quotient map and instead maintains derivation property of the $\iota_v$. On the other hand, for the symmetric tensors $Sym \subset T(V^*)$ one keeps the product compatible via the projections to the product on $S(V^*)$, but the $\iota_v$ induced from the tensor algebra are not derivations of this product on $Sym$.

P.S.: A somewhat similar situation appears, when one wants to define the contractions $\iota_v$ on $\Lambda(V^*)$ for $v\in V$: One can

a) define contractions $\iota_v$ on $\Lambda(V^*)$ which correspond under the projection map to the contractions on $Alt$.

b) define contractions $\iota_v$ on $\Lambda(V^*)$ such that these are graded derivations for the wedge product.

Also here the two requirements are incompatible. In this case, if one chooses option b), then the contraction $\iota_v$ is also adjoint to $v\wedge \cdot$ with respect to the natural pairing between $\Lambda(V)$ and $\Lambda(V^*)$, which is further discussed here: "Natural" pairings between exterior powers of a vector space and its dual

Answer 5

Question: "as I have tried to explain briefly. Said differently, it is natural to ask that the identification Sym2V→∼S2V should be the restriction of the projection map p:V⊗V→S2V. (Same story for the wedge product)."

Answer: If $V \cong k^n$ where $k$ is a field of characteristic zero, there is a canonical decomposition

$$V\otimes_k V \cong S^2(V) \oplus \wedge^2 V \oplus W$$

as $SL(V)$-modules. More generally

$$T^n(V):=V^{\otimes_k n} \cong \oplus_i V(\lambda_i)$$

is canonically the direct sum of a finite set of irreducible $SL(V)$-modules $V(\lambda_i)$. Given any Schur functor $\mathbb{S}_{\lambda}$ it is an open problem to determine the decomposition

$$\mathbb{S}_{\lambda}(V) \cong \oplus_i V(\lambda_i)$$

into a direct sum of irreducible $SL(V)$-modules.

Comment: "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)."

Answer: "Naturality" of the maps above in this context should be interpreted as "functorial". As an example: if $\phi: V \rightarrow W$ is a map of $k$-vector spaces there are many choices of isomorphisms

$$\eta: S^d(V^*) \cong S^d(V)^*,$$

(there are $k^{\binom{d+n-1}{n-1}}$ choices) but there is "one canonical choice" that is functorial: Define

$$\eta(\phi_1 \cdots \phi_d)(v_1\cdots v_d):= \sum_{\sigma \in S_d}\phi_{\sigma(1)}(v_1)\cdots \phi_{\sigma(d)}(v_d).$$

The diagram

$\require{AMScd}$ \begin{CD} S^d(W^*) @>>> S^d(W)^*\\ @V V V @VV V\\ S^d(V^*) @>>> S^d(V)^* \end{CD}

will commute for all maps $\phi$. If $V,W$ are $G$-modules for some group $G$ it happens to be that this map is $G$-linear.

Comment: "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?"

Answer: "A consistent choice" of maps should be interpreted as "a functorial choice of maps": A choice such that the following diagram commutes

$\require{AMScd}$ \begin{CD} Sym^d(V) @>>> S^d(V)\\ @V V V @VV V\\ Sym^d(W) @>>> S^d(W) \end{CD}

for all maps $\phi: V \rightarrow W$ - you must check if this is compatible with your "equalities" in 1 (and 2).If

Example: If $u,v\in V$ it follows the map

$$\rho: Sym^2(V) \rightarrow S^2(V)$$

is the following:

$$dudv:=u\otimes v + v\otimes u \rightarrow 2(u\otimes v) \in S^2(V):=V\otimes V/k\{u\otimes v -v\otimes u\}.$$

Hence $\rho(dudv)=2(u\otimes v):=2dudv \in S^2(V)$. Hence the "natural map" multiplies by $2$. If you choose $dudv:=\frac{1}{2}(u\otimes v+ v \otimes u)$ it follows $\rho(dudv)=dudv \in S^2(V)$. Hence it seems the "natural choice" is this last one.

There is something called "divided powers" that is related:

https://en.wikipedia.org/wiki/Divided_power_structure