idea of the star position in pullback, pushforward notation

Solution 1:

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.

Solution 2:

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 $\alpha=\phi^*\beta$.