I'm studying differential geometry using the book "Godinho Natàrio - An introduction to Riemannian Geometry". These are the definitions and theorems I'm working with:

Definition 1 (Pullback of a linear map) Let $V,W$ be finite dimensional real vector spaces, $F : V → W$ be a linear map. Then for every $k$ positive integer we define the pullback of $F$ as $$ F^* : \mathcal{T}^k(W^*) \to \mathcal{T}^k(V^*) \quad \quad (F^*T)(v_1, \dots, v_k) = T(F(v_1), \dots, F(v_k)) $$ for any $v_1, \dots, v_k \in V$. Here $\mathcal{T}^k(W^*)$ is the space of $k$-covariant tensors on $W$.

Theorem 1.13 Let $V$ be a $n$-dimensional real vector space, $F : V → V$ be a linear map and let $T \in 􏰁\Lambda^n(V^*)$ (the space of $n$-covariant alternating tensors on $V$). Then $F^*T = (\det A)T$ , where $A$ is any matrix representing $F$ .

Definition 2(Pullback of a tensor field) Let $M, N$ be smooth manifolds, $f : M \to N$ be a differentiable map. Then, each differentiable $k$-covariant tensor field $T$ on $N$ defines a $k$-covariant tensor field $f^*T$ on $M$ in the following way: $$ (f^*T)_p(v_1,...,v_k) = T_{f(p)}((df)_p(v_1),...,(df)_p(v_k)) $$ for any $ v_1, \dots,v_k \in T_pM$.

Then this last definition applies also to a differential form (being it a $k$-covariant differentiable alternating tensor field).

What I can't understand is the following remark at page 73 :

Let $M,N$ be smooth manifolds and let $f: M \to N$ be a differentiable map s.t. $\dim(M)=\dim(N)=n$. Let $p \in M$ and consider a coordinate systems $x = (x^1, \dots, x^n)$ around $p$ s.t. $x: V \to \mathbb{R}^n$ and $y=(y^1, \dots, y^n)$ around $f(p)$ s.t. $y: W \to \mathbb{R}^n$. Let $\hat{f}:= y \circ f \circ x^{-1}$ be the local representation of $f$. Then from Theorem 1.13 : $$(f^*(dy^1 \wedge \dots \wedge dy^n))_p = \det(d \hat{f})_{x(p)}(dx^1\wedge \dots \wedge dx^n)_p$$

How can I apply Theorem 1.13 in this situation? I mean, "translating" Definition 2 into Definition 1, I have the pullback of the linear map $dF_p : T_pM \to T_{f(p)}N$ applied to the element $dy^1 \wedge \dots \wedge dy^n$ of $\Lambda^n(T_{f(p)}N^*)$. But those vector spaces are not the same (as in the hypothesis of the Theorem)!


I think it can be fixed in this way:

Let $I_1 : \mathbb{R}^n \to T_pM$ and $I_2 : \mathbb{R}^n \to T_{f(p)}N$ be two isomorphisms s.t.: $$ I_1(e^i) = \frac{\partial}{\partial x^i} \quad \quad I_2(e^i) = \frac{\partial}{\partial y^i} \quad \forall \, i = 1, \dots, n $$ where $ \{ e^1, \dots, e^n \}$ is the standard basis of $\mathbb{R}^n$.

Then $$ F := I_2^{-1} \circ df_p \circ I_1 : \mathbb{R}^n \to \mathbb{R}^n$$ is an endomorphism in $\mathbb{R}^n$. By Theorem 1.13

$F^* = \det(A) \cdot$ being $A$ the matrix representing $F$.

By pullback's properties we have

$$\det(A) \cdot = F^* = (I_2^{-1} \circ df_p \circ I_1)^* = I_1^* \circ (df_p^*) \circ (I_2^*)^{-1} \Rightarrow df_p^* = (I_1^*)^{-1} \circ (\det(A) \cdot) \circ I_2^*$$


$$ (f^*(dy^1 \wedge \dots \wedge dy^n))_p = df_p^* (dy^1 \wedge \dots \wedge dy^n)= ((I_1^*)^{-1} \circ (\det(A) \cdot) \circ I_2^*) (dy^1 \wedge \dots \wedge dy^n)= \det(A) (I_2 \circ I_1^{-1})^*(dy^1 \wedge \dots \wedge dy^n) $$

Moreover $$I_2 \circ I_1^{-1} : T_p(M) \to T_{f(p)}N \quad \quad \frac{\partial}{\partial x^i} \mapsto \frac{\partial}{\partial y^i} $$ and then

$$(I_2 \circ I_1^{-1})^*(dy^i) \biggl (\frac{\partial}{\partial x^j} \biggr) = dy^i \biggl ( (I_2 \circ I_1^{-1}) \biggl (\frac{\partial}{\partial x^j} \biggr) \biggr ) = dy^i \biggl ( \frac{\partial}{\partial y^j} \biggr ) = \delta_{ij} = dx^i \biggl (\frac{\partial}{\partial x^j} \biggr) $$ i.e. $$(I_2 \circ I_1^{-1})^*(dy^i) =dx^i $$ and then since $ (I_2 \circ I_1^{-1})^* ( dy^i \wedge dy^j) = ((I_2 \circ I_1^{-1})^*dy^i) \wedge ((I_2 \circ I_1^{-1})^*dy^j)$ we have

$$(f^*(dy^1 \wedge \dots \wedge dy^n))_p = \det(A)(dx^1 \wedge \dots dx^n) $$

and it is easy to show that $A = [d\hat{f}]_{ij}$.

Is it ok?


You can skip the $\mathbb{R}^n$ inbetween. Since $M$ and $N$ have the same dimension, their tangent spaces are isomorphic, although there is no canonical choice of isomorphism. Once you have chosen charts on both manifolds you get an isomorphism. The determinant of the Jacobian matrix though does not depend on that choice.

  • $\begingroup$ But if I want to apply Theorem 1.13 I need an endomorphism. In the way (I think) you are suggesting I don’t have an endomorphism. Do you think my proof is correct? $\endgroup$
    – Bremen000
    Oct 25 '18 at 7:37
  • $\begingroup$ Your proof seems to be correct. I'm just saying that you can skip the $\mathbb{R}^n$ step in-between. In other words, one can prove Thm. 1.13 for $F:V\to W$ for $V,W$ two $n$-dimensional real vector spaces. After choosing bases $e_i$ and $f_i$ on both $V$ and $W$, let $\mathcal{I}:V\to W$ be the linear map that sends each $e_i$ to $f_i$. You get an endomorphism of $V$ by $\mathcal{I}^{-1}\circ F$, to which Thm. 1.13 can be applied. One only needs to check that the result does not depend on the choice of bases. Then apply this generalization to the differential of a map between manifolds. $\endgroup$
    – S.Surace
    Oct 25 '18 at 8:04

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