Pushforward of covariant and contravariant tensor.

Question

Le $F : M \rightarrow N$ be a map between manifolds. What is the pushforward of a covariant or contravariant tensor? I think that for a covariant tensor $T : T_pM \times...\times T_pM \rightarrow \mathbb{R}$ it is $F_*T(X_1,...,X_n)=T(F_*X_1,...,F_*X_n)$ but what for a contravariant tensor $T : T_p^*M \times...\times T_p^*M \rightarrow \mathbb{R}$? (note here $X_1,...,X_n$ are derivation of $T_pN$).

Answer 1

There seems to be some confusion of terminology here. In summary, given a smooth map $F: M \to N$ between manifolds:

• The differential of $F$ defines at each point $p \in M$ a pushforward $(F_*)_p : T_p M \to T_{F(p)} N$. (In general, the pushforward $F_* X$ of a vector-field on $M$ is not well defined.)
• If $T : T_p M \times \cdots \times T_p M \to \Bbb R$ is a (covariant) tensor, in general there is no such thing as a pushforward $T_* F$ on $N$. Instead, if $S : TN \times \cdots \times TN \to \Bbb R$ is a (covariant) tensor on $N$, then $F$ determines a pullback $F^* S : TM \times \cdots \times TM \to \Bbb R$ of $N$ to $M$ defined by $$(F^* S)_p (X_1, \ldots, X_k) = S_{F(p)} (F_* X_1, \ldots, F_* X_k)$$ (I suggest convincing yourself that both sides make sense).
• If $T : T^*_p M \times \cdots \times T^*_p M \to \Bbb R$ is a contravariant tensor, by dualizing we can identify it with an element of of $T_p M \times \cdots \times T_p M$. If $T$ is simple, so that we can write it as a sum of simple tensors $\alpha_1 \otimes \cdots \otimes \alpha_k$, $\alpha_i \in T_p M$, (it is rare to use this construction but) we can define the pushforward by $$F_* T := (F_* \alpha_1) \otimes \cdots \otimes (F_* \alpha_k) .$$ Then, we can define $F_* T$ for arbitrary contravariant tensors by linearity. As for vector fields, in general there is no such thing as a pushforward of a contravariant tensor field on $M$.