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i would like to know if there is some idea behind the position of the star in the pullback, pushforward notation or if it is just some notation without background?

What's the reason for the star to be up at the pullback and not down?*

I've read that there is no real standard notation (Spacetime and Geometry, Sean Carroll) and i wanted to find out what the original idea behind the notation was. I didn't find anything useful, so maybe you know?

Edit: *

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  • $\begingroup$ I'm not sure there is an "idea" lurking behind the notation. What context are you talking about, precisely? It sounds like you're talking about a linear function and its induced map on things like tensors? The natural generality for this notion, $f$ to $f_*$ or $f^*$ would be when you are only considering one functor between two fixed categories. Basically, it's just minimal notation that you need to keep track of what you're doing when you're in that kind of context. $\endgroup$ – Ryan Budney Sep 30 '11 at 14:42
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    $\begingroup$ One can trace the choices to where indices are put in the Ricci notation for tensors, going all the way to the beginning of time. $\endgroup$ – Mariano Suárez-Álvarez Sep 30 '11 at 14:44
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    $\begingroup$ @RyanBudney: My first motivation was, that i always forget where i should put the star and i thought maybe there is some way i should think about the starposition. Like the indices of vectors and 1-forms in physic abstract indices notation. I'm especially interested in the context of how it is used in general relativity (since i'm physics student). I thought maybe there is some connection to the indices-positions of the objects it is used on (vectors, 1-forms). $\endgroup$ – ungerade Sep 30 '11 at 15:00
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    $\begingroup$ @ungerade: An upper index on a tensor is ‘contravariant’, as is the operation of pullback—at least category-theoretically. That is what Mariano has been trying to say. $\endgroup$ – Zhen Lin Sep 30 '11 at 16:05
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    $\begingroup$ My way of remembering the notation was by thinking that $f_*$ is kicking a vector field forward, whereas $f^*$ is pulling a form back by the hair. $\endgroup$ – Sam Lisi Oct 4 '11 at 16:33
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Well, I don't really know the reason why the stars are where they are. But here's how I remember it:

  1. You can push-foward the the tangent space for any morphism of smooth manifolds, and you can pull-back the cotangent space.
  2. The cotangent bundle is usually written using the star for dual as $T^*M$.
  3. Ergo, pull-backs by smooth maps are $\phi^*$.

Note that it is a common convention that the "upper" starred objects map things in the "reverse" direction. $\phi: M\to N$ and $\phi^*: TN\to TM$ for smooth manifolds. $A: V\to W$ a linear map between inner-product spaces, the adjoint $A^*:W\to V$ goes the other way.


If I were to engage in some baseless speculation:

In the context of differentiable manifolds (where this question seems to be motivated), the induced (by a smooth map between manifolds) mapping of the cotangent bundle is the first one that you really need a new notation for. For the tangent bundle, if $\phi: M\to N$ you can just use $d\phi: TM\to TN$. So to me it is quite plausible that someone needed a notation for the induced mapping of the cotangent bundle, and decided to adorn $\phi$ with a star (for reasons unknown). Perhaps there is already an established tradition of adding this type of symbols in the superscript position (perhaps to allow enumerating several maps $\phi^*_1, \phi^*_2, \ldots$). And when it comes time where a symbol for the pushforward is needed, it is actually quite natural to just move the star down as it were.

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This may be too late but I have the following observation. If $\phi:M\to N$ and $\alpha = \phi^*\beta$, then the transformation law for covectors in coordinates is $$ \alpha_k = \sum_i \frac{\partial y_i}{\partial x_k}\beta_i , $$ where $y=\phi(x)$. Denoting by $D\phi$ the matrix representing the derivative of $\phi$, and thinking of $\alpha$ and $\beta$ as row vectors, the above transformation law in matrix notation reads $$ \alpha = \beta D\phi . $$ Clearly, if we think of $\alpha$ and $\beta$ as column vectors, then we have $$ \alpha = (D\phi)^T\beta . $$ At least visually, it looks very close to $\alpha=\phi^*\beta$.

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