Does the wedge product of bundle-valued forms induce a universal object?

Given a smooth manifold $M$ and a vector bundle $E$ over $M$, the $C^\infty(M)$-module of $E$-valued $p$-forms on $M$ is defined to be $$\Omega^p(M; E) := \Gamma_M\left( \bigwedge^p T^*M \otimes E \right).$$

The wedge product of a $E_1$-valued $p$-form $\omega_1 \in \Omega^p(M; E_1)$ with a $E_2$-valued $q$-form $\omega_2 \in \Omega^q(M;E_2)$ is defined in R. W. Sharpe's Cartan geometry text to be the $E_1 \otimes E_2$-valued $p+q$-form $\omega_1 \wedge \omega_2 \in \Omega^{p+q}(M; E_1 \otimes E_2)$ given by $$(\omega_1 \wedge \omega_2) (v_1, \ldots v_{p+q})=\sum_{\text{(p,q) shuffles \sigma}} (-1)^{\text{sgn}(\sigma)} \omega_1(v_{\sigma(1)}, \ldots, v_{\sigma(p)})\otimes \omega_2(v_{\sigma(p+1)}, \ldots, v_{\sigma(p+q)}).$$

I am wondering if this multiplication induces a universal object of some sort?

For example, the wedge product of $\mathbb{R}$-valued forms gives you the exterior algebra $\Omega^{\bullet}(M)$, which is the free graded-commutative algebra on $\Omega^1(M)$.

This is more generally true for the exterior algebra $\bigwedge^\bullet V$ for any module $V$ over a commutative ring. Likewise, the tensor algebra $\bigotimes^\bullet V$ is the free algebra on $V$, and the symmetric algebra $\bigodot^\bullet V$ is the free commutative algebra on $V$.

Does the wedge product of bundle-valued forms induce a universal object of some sort as well?

Also, I would really appreciate a reference on bundle-valued forms, especially one from a categorical viewpoint. I haven't been able to find a comprehensive treatment yet.

• I recall Morita's book The Geometry of Differential Forms is a reference for bundle-valued forms, but it's been a while and I don't think it includes a categorical perspective.
– Neal
Commented May 11, 2015 at 12:34
• General remark: Don't just look at global sections, look at the whole sheaf. Commented May 12, 2015 at 13:14
• @Martin Brandenburg Could you explain how looking at the whole sheaf helps here?
– ಠ_ಠ
Commented May 12, 2015 at 13:28
• For any one who wish to understand bundle-valued forms, as said by Neal, it is in The geometry of Differential forms by Morita.. it is in page number 191.. you can also see it in contents... chapter 5, section 5.3, subsection e, named Differential forms with valued in a vector bundle.....
– user537667
Commented Jul 25, 2019 at 15:51

This is quite an old question, but maybe I can say something useful for anyone who stumbles upon it.

The tldr of this answer is that you take the tensor product of your two differential forms and then you precompose with map from the bundle of $$n=p+q$$-forms to the tensor product of the bundle of $$p$$-forms and the bundle of $$q$$-forms. Most of what follows is an explanion of how that latter inclusion map is built in a general symmetric monoidal abelian category (aka a tensor category).

We work in an abelian category $$\mathcal{A}$$ with a symmetric monoidal product $$\otimes$$ that preserves finite direct sums in each variable.

For $$E\in\mathcal{A}$$, the symmetric monoidal structure gives us an $$S_n$$ action on $$E^{\otimes n}$$, the $$n$$-fold tensor product of $$E$$ with itself. Since we're in an abelian category, each object also has an automorphism given by multiplication by $$-1$$ (more precisely, this is the negative of the identity morphism). This determines a $$C_2=\{1,-1\}$$ action.

Since $$\otimes$$ preserves direct sums, the $$S_n$$ action on $$E^{\otimes n}$$ commutes with the $$C_2$$ action and we have a total action given by a functor $$\phi_E:B(S_n\times C_2)\to\mathcal{A}$$ (where for a group $$G$$, $$BG$$ is the one-object category with endomorphism monoid $$G$$).

Note that the function $$\psi:S_n\times C_2\to C_2$$ given by $$q:(\sigma,\alpha)\mapsto\mathrm{sgn}(\sigma)\alpha$$ is a group homomorphism and so induces a functor $$\psi:B(S_n\times C_2)\to BC_2$$.

We define $$\bigwedge_n E:=\mathrm{Ran}_{\psi}\phi_E$$, the right Kan extension of $$\phi_E$$ along $$\psi$$. This is the subobject of $$E^{\otimes n}$$ on which the symmetric group acts by multiplication by the signs of its elements (permutations).

Let $$p+q=n$$ and write $$S_{p,q}:=S_p\times S_q$$. We have an inclusion $$\iota:S_{p,q}\to S_n$$, where a $$p$$-permutation and a $$q$$-permutation give us an $$n$$-permutation by taking the disjoint union of the two permutations (i.e. we never swap any of the first $$p$$ elements with any of the last $$q$$ elements of our set).

Taking $$\mathrm{Ran}_{\psi\circ\iota}(\phi\circ\iota)$$ gives us $$\bigwedge_p E\otimes \bigwedge_q E$$. Whiskering along $$\iota$$ and the universal property of this latter right Kan extension now give us a morphism $$\eta:\bigwedge_n E\to\bigwedge_p E\otimes\bigwedge_q E$$.

So finally, if we started out with morphisms $$f:\bigwedge_p E\to U$$ and $$g:\bigwedge_q E\to V$$, we obtain $$f\wedge g:\bigwedge_n E\stackrel{\eta}{\to}\bigwedge_pE\otimes\bigwedge_qE\stackrel{f\otimes g}{\to} U\otimes V$$

If you had a way of multiplying $$U$$ and $$V$$ together to get stuff in a fourth object $$W$$ (i.e. a map $$\kappa:U\otimes V\to W$$), then you can define $$f\wedge_{\kappa}g$$ to be $$\kappa\circ f\wedge g$$. One example of this that might be of interest would be the case where $$U=V$$ and $$U$$ carries an inner product structure $$U\otimes U\to \mathbb{R}$$ (if you're working over $$\mathbb{R}$$, say, as one typically does in differential topology) or perhaps an algebra structure $$U\otimes U\to U$$.