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?


2 Answers 2


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
    Commented Oct 25, 2018 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
    Commented Oct 25, 2018 at 8:04

This can be done on a more general context (m-manifold/n-manifold/non simple k-form), according to the Pullback section of Differential_form article of Wikipedia:

Let $φ : M \to N$ be smooth, and let $ω$ be a smooth k-form on $N$. Assume that $x^1, ..., x^m$ are coordinates on $M$, that $y^1, ..., y^n$ are coordinates on $N$, that $f$ is the representation of $φ$ under those coordinates, and that these coordinate systems are related by the formulas $y^i = f_i(x^1, ..., x^m)$.

Let $φ_i=f_i \circ x$ where $x$ is the chart map for $M$. $y$ is the chart map on $N$.

Make some concept clear: {$y^1, ...,y^n$} are smooth map to $R$ on $N$, {$dy^1, ..., dy^n$} are 1-forms on $N$, and at each point of $N$, the evaluation of $dy^1, ..., dy^n$ form a basis for any 1-forms on $N$ evaluated at that point (all evaluations are cotangent vector). Also at each point of $N$, the evaluation of all $dy^{i_1}\wedge...\wedge dy^{i_k}$ ($i_1\le...\le i_k$) form a basis for any k-form on $N$ evaluated at that point (all evaluations are kth exterior power of cotangent vector). These are also true for $dx$ and $M$. $dφ_i$ are also 1-forms on $M$.

Then $ω$ can be expanded with $dy^{i_1}\wedge...\wedge dy^{i_k}$ point-wisely on $N$:

$$ω=\sum_{i_1<...<i_k}w_{i_1...i_k}dy^{i_1}\wedge...\wedge dy^{i_k}$$

here $w_{i_1...i_k}$ is a smooth map from $N$ to $R$ which is also a 0-form on $N$.

your question is the special case when $m=n=k$ and the only $w_{i_1...i_k}=w_{i_1...i_n}=1$.

Here we continue with the general case, then:

$$φ^*ω=φ^*(\sum_{i_1<...<i_k}w_{i_1...i_k}dy^{i_1}\wedge...\wedge dy^{i_k}) \\=\sum_{i_1<...<i_k}φ^*(w_{i_1...i_k}\wedge dy^{i_1}\wedge...\wedge dy^{i_k}) \\=\sum_{i_1<...<i_k}φ^*(w_{i_1...i_k})\wedge φ^*(dy^{i_1})\wedge...\wedge φ^*(dy^{i_k}) \\=\sum_{i_1<...<i_k}(w_{i_1...i_k}\circ φ)\wedge dφ_{i_1}\wedge...\wedge dφ_{i_k} \\=\sum_{i_1<...<i_k}(w_{i_1...i_k}\circ φ) dφ_{i_1}\wedge...\wedge dφ_{i_k} $$

In the above result, since each: $$ dφ_{i_1}\wedge...\wedge dφ_{i_k}=(\sum_{j=1}^{m}(\frac{\partial f_{i_1}}{\partial x^j}\circ x)dx^j)\wedge...\wedge(\sum_{j=1}^{m}(\frac{\partial f_{i_k}}{\partial x^j}\circ x)dx^j) \\=\sum_{j_1<...<j_k}\{coefficient\}dx^{j_1}\wedge...\wedge dx^{j_k} $$

The $\{coefficient\}$ for $dx^{j_1}\wedge...\wedge dx^{j_k}$ will be the sum of every "permutation factor" for each permutation of $j_1,...,j_k$. The "permutation factor" for each permutation $σ$ of $j_1,...,j_k$ is formed by picking the expansion coefficient of $dx^{σ(j_z)}$ for $dφ_{i_z}$ then multiply the $k$ coefficients together with the permutation sign for permutation of wedge product.


$$ \{coefficient\ for\ dx^{j_1}\wedge...\wedge dx^{j_k}\} \\=\sum_{σ \in S(j_1,...,j_k)}\prod_{z=1}^{k}sgn(σ)\frac{\partial f_{i_z}}{\partial x^{σ(j_z)}}\circ x \\=(\sum_{σ \in S_k}\prod_{z=1}^{k}sgn(σ)\frac{\partial f_{i_z}}{\partial x^{j_{σ(z)}}})\circ x $$

This (in the parenthese) is the determinant of the $k*k$ matrix whose elements are $a_{pq}=\frac{\partial f_{i_p}}{\partial x^{j_q}}$, which is written as $$\frac{\partial(f_{i_1},...,f_{i_k})}{\partial(x^{j_1},...,x^{j_k})}$$

Final result:

$$ φ^*(\sum_{i_1<...<i_k}w_{i_1...i_k}dy^{i_1}\wedge...\wedge dy^{i_k}) \\=\sum_{i_1<...<i_k}(w_{i_1...i_k}\circ φ) \sum_{j_1<...<j_k}(\frac{\partial(f_{i_1},...,f_{i_k})}{\partial(x^{j_1},...,x^{j_k})}\circ x)dx^{j_1}\wedge...\wedge dx^{j_k} $$

Then come back to the special case in the question:

$$φ^*(dy^1\wedge...\wedge dy^n) =(\frac{\partial(f_1,...,f_n)}{\partial(x^1,...,x^n)}\circ x)dx^1\wedge...\wedge dx^n $$

, which is the $\det(d \hat{f})_{x(p)}$ in the question.

Note: althougn it won't make it ambiguous to ommit the chart map $x$ and map $φ_i$ here, but strictly speaking the result is a k-form on $M$ whose expansion coefficient should be smooth map from $M$ to $R$, not from $R^m$ to $R$. The coefficient without chart map $x$ as in the question is indeed the expansion coefficient for the pullback k-form on $R^m$ (if you also view $x^i$ as smooth functions on $R^m$ and $dx^i$ as 1-forms on $R^m$)


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