My understanding of differential forms is that, given the map(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \phi : V \to V^{*} [/tex]

If [tex] f \in V^{*} [/tex] is a 0-form (a real-valued function on [tex] V^{*} [/tex]), then [tex] \phi [/tex] defines the function [tex] \phi^{*} f [/tex] on [tex] V [/tex]

where [tex] \phi^{*}: F^{0}(V^{*}) \to F^{0}(V) [/tex]

as the function such that [tex] \phi^{*}f(p)=f(\phi(p)) [/tex].

So, since differential forms transform contravariantly, the 0-form (function) f transforms contravariantly (and therefore is contravariant.....).

Because [tex] f [/tex] doesn't have any indices, I decided to look online and see if that fact makes a difference in requiring the label "contravariant."

One of the first links was to a thread on physicsforums

https://www.physicsforums.com/showthread.php?t=311174

The debate is between "dx" and "MarkParis." It doesn't seem to reach an agreement in that thread, so I was wondering if anyone else could shed some light on it.

*IMO, the posts by "dx" are correct and meet my understanding of the topic. However, I'm still learning differential forms and there is certainly going to be nuances of the subject that I haven't understood yet....so I figured I'd see if anyone here has any input on this and/or the posts made in the linked thread.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Is a function contravariant?

**Physics Forums | Science Articles, Homework Help, Discussion**