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Spivak's "Calculus on Manifolds" defines the wedge product between a $k$-form and a $l$-form (alternated tensors $\in \Omega^k(V), \in \Omega^l(V)$ respectively as:

$$\omega(v_1,\cdots,v_k)\wedge \eta(v_{k+1},\cdots, v_{k+l})=\frac{(k+l)!}{k!l!} \frac1{k!} \sum\text{signed permutations} $$

I'm writing "signed permutations" to mean, for example with two one-forms:

$$ \omega(v_1)\wedge \eta(v_2) = \frac{(1+1)!}{1!1!}\frac1{1!}\omega(v_1) \eta(v_2) - \omega(v_2) \eta(v_1) $$

Anyway. Arnold's "Mathematical Methods of Classical Mechanics" defines the wedge product without the constant -- just the sum of signed permutations; so his product of two one-forms is half of Spivak's.

Is this a physics/pure maths difference in convention (so the constant factor doesn't matter in physics), or is this factor usually accounted for elsewhere in different expositions of the theory (for example, in exterior differentiation or symplectic forms, etc.)?

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    $\begingroup$ Are you sure that Spivak's definition is not $\frac{(k+l)!}{k! l!} \operatorname{Alt}()$ which amounts to the factor $\frac{1}{k! l!} \sum$ and not what you wrote? And are you sure that Arnold doesn't define it to be $\frac{1}{(k+l)!} \sum$? Those are the two popular conventions and from what I can tell, it isn't a math/physics thing but an author preference regarding whether you want to eliminate certain constants from some equations rather than from others. It is important to understand that not all choices of constant are possible (some would ruin associativity of the product which is bad) $\endgroup$
    – levap
    Feb 6, 2017 at 13:34
  • $\begingroup$ The formula for $\textrm{Alt}(T)$, where $T$ is an $m$-tensor is $\textrm{Alt}(T) := \frac{1}{m!} \sum_{\sigma \in S_m} (\textrm{sgn} \sigma) ({}^{\sigma}T)$. In the formula for the wedge product (as Spivak defines it), we have $T = \omega \wedge \eta$, a $(k + l)$-tensor, so $\omega \wedge \eta = \frac{(k + l)!}{k! l!} \textrm{Alt} (\omega \otimes \eta) = \frac{1}{k! l!} \sum_{\sigma \in S_{k + l}} (\textrm{sgn} \sigma) ({}^{\sigma}(\omega \otimes \eta))$. $\endgroup$ Feb 6, 2017 at 13:37
  • $\begingroup$ This discussion and this one might help. $\endgroup$
    – Ivo Terek
    May 14, 2018 at 23:13

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