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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.

hm2020
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Seub
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    This discussion over at MathOverflow seems pertinent, particularly from the standpoint of differential geometry: http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product – Branimir Ćaćić Aug 25 '14 at 18:21
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    On the algebraic side of things, it should be said that the projector onto symmetric $2$-tensors is somewhat special to linear algebra over a field of characteristic $\neq 2$. In particular, if you're just working with, say, Abelian groups *qua* $\mathbb{Z}$-modules, and you don't have division by $2$ at your disposal, then you really have to use the convention without the $1/2$'s. Of course, this is also a situation where $T^2(V)$ *won't* necessarily be a direct sum of $S^2(V)$ and $\wedge^2(V)$. – Branimir Ćaćić Aug 25 '14 at 18:29
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    Yes, thank you for mentioning that link, I should do it in my question maybe. But I don't agree with the accepted answer there. Basically, it saying that from the algebraic point of view, one choice is not more natural than the other. I think differently, as I have tried to explain in my question. – Seub Aug 25 '14 at 18:33
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    I think you've hit the nail on the head. For symmetric products, there's only one reasonable choice. But for wedge products, the convention that's most natural algebraically and the convention that's most natural for differential geometry are different. We just have to live with it. – Jack Lee Aug 25 '14 at 19:00
  • Sorry, it's not clear to me why you dropped the factor of $1/2$ for your differential geometry definition of the wedge. Doesn't the usual definition in that context also have the factor of $1/2$? – Muphrid Aug 25 '14 at 19:01
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    @Muphrid I dropped the 1/2 because in differential geometry, you want $dx \wedge dy$ to be the area form (determinant) on $\mathbb{R}^2$. Said differently, integrating a function times $dx \wedge dy$ should give you the Lebesgue integral, at least I think this is the convention everyone uses. This gives you no choice: $dx \wedge dy$ should be $\mathrm{det} = dx \otimes dy - dy \otimes dx$. – Seub Aug 25 '14 at 19:14
  • Odd. I come from a clifford algebra background; the geometric product used there should not have any scale factors in its relation to the tensor product, but the definition of wedge used there has the factor of $1/2$. Can you show that this definition without the $1/2$ gives the correct determinant? – Muphrid Aug 25 '14 at 19:43
  • @Muphrid $(dx \otimes dy - dy \otimes dx)((a,b),(c,d)) = dx((a,b))dy((c,d)) - dy((a,b))dx((c,d)) = ad - bc$. I had to do a double take too, though, because I'm so used to the $1/2$ as well. – Branimir Ćaćić Aug 25 '14 at 23:53
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    Yes, if you want $\operatorname{det} = e^1 \wedge \cdots \wedge e^n$ for $\{e^k\}$ the dual basis to the standard ordered basis on $\mathbb{R}^n$, then you shouldn't divide by $n!$. So, I guess you really just have to pick your convention based on what is fundamental in your own line of work. – Branimir Ćaćić Aug 26 '14 at 10:06
  • @JackLee Thanks for your comment, it might be as good an answer as I can hope for. – Seub Aug 26 '14 at 17:27

3 Answers3

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

amine
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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. $\displaystyle \alpha\wedge\beta=Alt(\alpha\otimes\beta)$

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.

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

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.

Shuchang
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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

hm2020
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  • I think the "right" explanation of my comment is more straightforward: $S^2 V$ is *defined* as a quotient of $V \otimes V$, so there is a canonical projection. No need to decompose any $SL(V)$-modules. – Seub Oct 07 '21 at 16:36
  • You may write $V\otimes_k V \cong S^2(V)\oplus U$ where the decomposition is a decomposition as $SL(V)$-modules, and hence the quotient $S^2(V) \cong V\otimes_k V/U$ is a canonically an $SL(V)$-module quotient. You may also define the symmetric square $S^2(V) \subseteq V\otimes_k V$ as a sub-module. – hm2020 Oct 07 '21 at 16:38
  • There is an appendix in Fulton/Harris "Representation theory - a first course" where some constructions in multilinear algebra are outlined which is recommended. – hm2020 Oct 07 '21 at 16:41
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    Sure, you may write that and more, but there is a pretty good principle called Occam's razor, I'd recommend checking it out. – Seub Oct 07 '21 at 16:46
  • @Seub - this may not be interesting for diffgeometers, but in algebra one wants maps to be "natural". One wants a "natural isomorphism" $S^d(V^*) \cong S^d(V)^*$ that is intrinsic (or $SL(V)$-invariant) where $V^*$ denotes the dual vector space/$SL(V)$-module. This construction is outlined in the FH book above - it is not difficult. – hm2020 Oct 07 '21 at 17:18
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    I'm not saying it's difficult, I'm saying it's irrelevant. Algebraists do not like pointless convolutions any more than geometers. (The quotient map is as a natural as it gets, no need for decompositions and isomorphisms) – Seub Oct 07 '21 at 23:42