Just like a vector field is a (smooth) choice of a vector at each point, a 1-form is a (smooth) choice of a covector at each point. The bundle of covectors is is contravariant, meaning that induces a “pull-back” map . In fact, this doesn’t just work for 1-forms; it works for differential forms of any degree. The definition for -forms is the first display-set equation in the post above. Basically it defines the pull-back of a -form on (which should be a -form on ) by its value on tangent vectors on . The natural thing to do is push those vectors *forward* to vectors on , and evaluate the original -form on them there. This is a pretty common construction when dealing with dualities like 1-forms and vector fields.

In general, there’s not a “natural” map induced by a map ; the natural map goes the other way, just like there’s no natural pull-back on vectors induced by . *However*, if you equip both bundles with a metric — and if you’re always thinking of these manifolds as sitting inside regular -dimensional space you get a metric induced from that inclusion — then you can use that to change vectors into covectors and vice versa. In that case, you do get a push-forward on differential forms. The catch is, the choice of metric is a choice of how to measure lengths of tangent vectors, and angles between tangent vectors to the same point. That says a lot about the “shape” of your space, and puts you into the realm of differential *geometry*; everything I’m doing here is without a choice of metric, and is differential *topology*. That might explain the difference.

As for and , I’m dealing here with the sheaf of smooth sections of the bundles of forms and functions. That is, given an open subset we consider the algebra of smooth real-valued functions on and call this algebra . In a small enough , you can think of this as “smooth slices” of the product space . Similarly, is the space of smooth -forms defined on , which is a module over the algebra . The important thing here is that we’re not just considering one stalk over one point at a time, but all the stalks over all the points , and that the choice of a point in each stalk must be “smooth” as we vary .

]]>We know that . We also know that any antisymmetric binary operation on must factor through . That is, if we have then it must factor as

In this case, we’re interested in a binary “multiplication” from to itself, meaning we’re looking for a map . This will be the Hodge star in the case where and , so , so this works out in three-dimensional space.

]]>Thank you for this response. I found it very illuminating.

To clear up: yes, I was asking specifically about [working] mathematicians, as you correctly interpreted.

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