# Determinant of the derivative of a map between manifolds

Ok, suppose $M,N$ are Riemannian manifolds and $F:M\to N$ is a smooth map between them. In a book I have here they consider that $\dim M=m \geq \dim N=n$ and that $x\in M$ is a regular point such that the derivative $DF(x):T_x M\to T_{F(x)}N$ is surjective. With this conditions, note that $\ker DF(X)^\perp$ is isomorphic to $T_{F(x)}N$.

So far so good, the problem comes when they start to talk about the determinant of $DF(x)$. As far as I know, we talk about determinant in the context of matrices. So here started my trouble, for $DF(x)$ is a linear map between abstract spaces, there is no obvious way to make $DF(x)$ into a matrix.

After a search on Google, I got these two definitions, which I wanted to share here to be sure they are valid. For the first one, fortunately, I had some knowledge of differential forms, so I didn't got so confused when seeing it.

Definition 1: given any base $(v_1,\ldots,v_n)$ of $\ker DF(x)^\perp$, we define $$\det (DF(x)) = \frac{\omega(DF(x)v_1,\ldots,DF(x)v_n)}{\mu(v_1,\ldots,v_n)},$$ where $\omega$ is the volume form on $N$ and $\mu$ is the volume form on $M$.

The second definition was told me by a friend and I couldn't find something useful on the Internet about it.

Definition 2: let $(\varphi, U)$ be a orthonormal coordinate system on $x$ and $(\psi, V)$ a orthonormal coordinate system on $F(x)$. Then $\det (DF(x)) = \det \left(D\Phi(\varphi^{-1}(x))\right)$, where $\Phi=\psi^{-1}\circ F\circ\varphi$.

This second definition also makes sense, assuming that the determinant doesn't depend on the charts chosen. In fact, this second definition is better, because it reduces the problem to the computation of a determinant of some matrix, which is something familiar. Just to be clear, $\varphi$ is a map from $U\subset \mathbb{R}^m$ to $M$ and $\psi$ is a map from $V\subset \mathbb{R}^n$ to $N$.

I have more than one question, some of them probably you will find easy to answer (assuming you are used to Riemannian Geometry).

1) This is a terminology question. Instead saying $(\varphi,U)$ is an orthonormal coordinate system, could I say it is an orthonormal chart ? I'm asking this because looks like it's common to use the term charts, except when you are gonna say it is orthogonal or orthonormal. In this case I found most of people prefer to use it with the term coordinate system.

2) From my reading, $(\varphi,U)$ is an orthogonal (orthonormal) coordinate system (or chart?) when $\left(\frac{\partial}{\partial x_1}|_x,\ldots,\frac{\partial}{\partial x_m}|_x\right)$ is an orthogonal (orthonormal) base for $T_xM$. Is my understanding correct?

3) The definitions 1 and 2 are really equivalent? I got a little suspicious about using "orthonormal" in definition 2. Maybe it could be "orthogonal", I don't know.

PS: questions 1 and 2 are important but are technical details necessary to ask question 3. The last question is the most important of all.

• The derivative $DF(x) : T_x M \to T_{F(x)} N$ is a linear map between vector spaces $T_x M$ and $T_{F(x)} N$ by definition. So to speak of its matrix, all you have to do is pick a basis of the vector spaces! And then the determinant is the determinant of that matrix. – Jon Warneke Nov 7 '16 at 23:27
• These vector spaces can be very abstract. Just picture projective manifolds, quotient manifolds, level sets manifolds, etc. I don't think picking basis will always be so easy as you speak. Probably this is the reason why the OP is relying on volume forms and charts to work with. – Integral Nov 8 '16 at 0:37
• Definition 2 not always makes sense because existence of a orthonormal coordinate system (OCS) is equivalent to $\ g|_U \$ being flat (isometric to some open set of $\mathbb{R}^m$) and it is not always the case. Take $\ M = S^2 \$ for example and you do not have OCS in any point. – Gustavo Nov 8 '16 at 22:50

Both definitions you suggest do not compile:

1. The volume form $\omega$ on $M$ eats $m$ tangent vectors at each point. How do you feed it with only $n$ vectors?
2. The matrix $D\Phi(\varphi^{-1}(x))$ is an $n \times m$ matrix. How do you define the determinant of a non-block matrix?

In order to clarify things let me discuss first the linear algebra relevant for your question. The main point is that there is no notion of determinant for a linear map between two vector spaces of different dimensions and there is no notion of determinant for a linear map between two different vector spaces of the same dimension without a choice of an extra data.

If $T \colon V \rightarrow V$ is a linear operator on a finite dimensional vector space (so the domain and codomain are the same), you can define $\det(T)$ by choosing a basis $\mathcal{B}$ for $V$ and defining $\det(T) := \det([T]_{\mathcal{B}})$ where $[T]_{\mathcal{B}} \in M_n(\mathbb{F})$ is the square matrix that represents the operator $T$ with respect to the basis $\mathcal{B}$. This definition uses a basis but is, in fact, independent of the basis we work with as $$\det([T]_{\mathcal{B}'}) = \det(P^{-1} [T]_{\mathcal{B}} P) = \det([T]_{\mathcal{B}})$$ where $P$ is the change of basis matrix $P =[\operatorname{id}]_{\mathcal{B}}^{\mathcal{B'}}$.

If $T \colon V \rightarrow W$ is a linear map between two vector spaces of the same (finite) dimension, you can try and define $\det(T)$ by representing $T$ as a matrix. However, to represent $T$ as a matrix you need to pick two different bases $\mathcal{B}$ for $V$ and $\mathcal{C}$ for $W$ and if $\mathcal{B}', \mathcal{C}'$ are other bases, we have

$$[T]_{\mathcal{C}}^{\mathcal{B}} = [\operatorname{id}]^{\mathcal{C}'}_{\mathcal{C}}[T]^{\mathcal{B}'}_{\mathcal{C}'} [\operatorname{id}]^{\mathcal{B}}_{\mathcal{B}'}$$

and so there's no reason that $\det([T]_{\mathcal{C}}^{\mathcal{B}}) = \det([T]^{\mathcal{B}'}_{\mathcal{C}'})$.

To save the situation, we should go back to the case $T \colon V \rightarrow V$ and reinterpret $\det(T)$ differently. If $V = \mathbb{R}^n,$ the scalar $\det(T)$ is the signed factor by which $T$ scales the volume of an $n$-dimensional parallelotope. In an abstract vector space $V$, we have no natural notion of (signed) volume of an $n$-dimensional parallelotope. Such notion is provided by a choice of a volume form $0 \neq \omega_V \in \Lambda^{\text{top}}(V^{*})$. Given such a volume form, we can define $\det(T)$ as the unique scalar such that $T^{*}(\omega_V) = \det(T) \omega_V$ and the nice thing is that this definition is actually independent of the volume form!

Finally, to define a notion of a determinant for a map $T \colon V \rightarrow W$, we equip $V$ and $W$ with volume forms $\omega_V \in \Lambda^{\text{top}}(V^{*})$ and $\omega_W \in \Lambda^{\text{top}}(W^{*})$ and define $\det(T)$ by the equation $T^{*}(\omega_W) = \det(T) \omega_V$. This definition depends on both $\omega_V$ and $\omega_W$.

Now assume $(V, g_V)$ is an inner product space. Even though you provided extra data, there is still no natural volume form that is defined on $V$. In order to define a natural form, you need to choose an orientation for $V$ (and equivalence class $\mathfrak{o}_V$ of elements of $\Lambda^{\operatorname{top}}(V)$ or $\Lambda^{\operatorname{top}}(V^{*})$ depending on your definitions). Once chosen, there is a unique volume form $\omega_{g_V,\mathfrak{o}_V}$ that behaves nicely with the metric and the orientation (this is precisely the Riemannian volume form). It is determined by the fact that if $(e_1, \dots, e_n)$ is a positive orthonormal basis of $V$ then $\omega_{g_V,\mathfrak{o}_V}(e_1 \wedge \dots \wedge e_n) = 1$.

In your situation you have a surjective linear map $T \colon (V, g_V, \mathfrak{o}_V,\omega_V) \rightarrow (W, g_W, \mathfrak{o}_V,\omega_W)$ with $\dim V = m, \dim W = n$. The map $T|_{(\ker T)^{\perp}} \colon (\ker T)^{\perp} \rightarrow W$ is a linear map between two vector spaces of the same dimension and so we can try and make sense of $\det \left( T|_{(\ker T)^{\perp}} \right)$. The right hand side has by assumption a volume form but the left hand side is only a subspace of a space that has a volume form. In general, a subspace of a space with a volume form doesn't get a volume form but $(\ker L)^{\perp}$ has an inner product (the restriction of $g_V$) and we can give it an orientation using the map $T$ and these two structures endow $(\ker T)^{\perp}$ with an orientation and allow us to talk of the determinant.

Finally, we can provide corrected versions of your definitions for $\det T$:

1. If $v_1, \dots, v_n$ is a basis of $( \ker T)^{\perp}$ such that $Tv_1, \dots, Tv_n$ is a positive basis of $W$, complete it to a positive basis $v_1, \dots, v_n, u_1, \dots, u_{m-n}$ of $V$ and then $$\det \left( T|_{(\ker T)^{\perp}} \right) = \frac{\omega_W(Tv_1, \dots, Tv_n)}{\omega_V(v_1, \dots, v_n, u_1, \dots, u_{m-n})}.$$
2. If $\mathcal{B} = (v_1, \dots, v_n)$ is an orthonormal basis of $( \ker T)^{\perp}$ such that $Tv_1, \dots, Tv_n$ is a positive basis of $W$ and $\mathcal{C} = (w_1, \dots, w_m)$ is a positive orthonormal basis of $W$ then $\det \left( T|_{(\ker T)^{\perp}} \right) = \det\left(\left[ T|_{(\ker T)^{\perp}} \right]_{\mathcal{C}}^{\mathcal{B}}\right)$.

I'll leave it to you to verify that the definitions are equivalent and consistent with what I described before.

• Thank you very much, you were very helpful. I still have two questions. 1) Is it possible to drop the orientation hypothesis and use the absolute value of the volume forms? 2) About your second question for me, I understand my (basic) error. In particular, in the definition with charts can I use $\det(D\Phi(\varphi^{-1}(x))|_{\ker D\Phi(\varphi^{-1}(x))^\perp})$ ? If this is correct, is enough to work with orthogonal (or orthonomal) charts to guarantee invariance of the determinant? Thanks. – diff_math Nov 9 '16 at 19:46
• @diff_math: Regarding your first question, there is no volume form if you don't have an orientation! Any volume form determines an orientation by declaring that $(v_1, \dots, v_n)$ is positively oriented if and only if $\omega(v_1, \dots, v_n) = \omega(v_1 \wedge \dots \wedge v_n) > 0$. If you don't want to use orientation, you can work with densities instead and then there is a notion of "determinant" which is the absolute value of the regular determinant and this doesn't depend on orientation. Riemannian manifolds don't have a canonical volume form (for this, the manifold needs to be – levap Nov 9 '16 at 22:03
• oriented and you need to pick an orientation) but they have a canonical density. Using densities pretty much amounts to putting an absolute value on the volume forms. Regarding the second question, you can pick orthonormal charts in such a way that the $n \times m$ matrix representing $D\Phi(\varphi^{-1}(x))$ will consist of an $n \times n$ matrix and a $n \times (m - n)$ zero matrix. The determinant of the $n \times n$ matrix will give you the determinant of $dF_{\varphi^{-1}(x)}|_{(\ker dF_{\varphi^{-1}(x)})^{\perp}}$ up to a sign (which is determined by the orientations). – levap Nov 9 '16 at 22:08
• Note that your "orthonormal coordinates" should be orthonormal only at the points $x,F(x)$. You can't get anything stronger because the existence of orthonormal coordinates which are orthonormal at each point means that yours manifolds are flat. – levap Nov 9 '16 at 22:11
• @diff_math: Sorry, I mean orthonormal. And it should read $\frac{\partial}{\partial y^i}|_{q}$ and not $\frac{\partial}{\partial y_i}|_{p}$. – levap Nov 12 '16 at 0:50

My bad. I totally misunderstood your original confusion, but I think I get it now (I've removed my previous posts). So your chart is a subset $U\subset M$ and a homeomorphism $\varphi$ to an open set of $\mathbb{R}^n$. To pinpoint the location of $\varphi(p)\in \mathbb{R}^n$, we need to express it in coordinates $(x_1,...,x_n)$. At each point of $U$, the coordinate directions have to span $\mathbb{R}^n\cong T_xM$. These are the $(\frac{\partial}{\partial x_1}|_x,...,\frac{\partial}{\partial x_n}|_x)$ from your 2nd question. When these directions are orthogonal (orthonormal), we say the coordinate chart is orthogonal (orthonormal).

• Perfect. Well, now I just need to know if definitions 1 and 2 really agree. thanks. – diff_math Nov 5 '16 at 14:32

This is not a complete answer, but just what I'm thinking, perhaps you could fill the gaps.

The differential of the map $\Phi=\psi^{-1}\circ F\circ \varphi$ is an $n\times m$ matrix. It is the product of the $n\times n$ matrix $D\psi^{-1}$, the $n\times m$ matrix $DF$, and the $m\times m$ matrix $D\varphi$. But this map vanishes on the $(m-n)-$dimensional subspace corresponding to the preimage of $\ker DF$ in $T_x\mathbb{R}^m\cong \mathbb{R}^m$. So the interesting stuff happens on an $\mathbb{R}^n$ sitting inside of $\mathbb{R}^m$ and we can view the composition as $$[D\Phi_|]_{n\times n}=[D\psi]^{-1}_{n\times n}[DF_|]_{n\times n}[D\varphi_|]_{n\times n}$$ and the determinant of this matrix makes sense. Then we can rewrite as $$[DF_|]=[D\psi][D\Phi_|][D\varphi_|]^{-1}$$ and we have $$\det(DF_|)=\frac{\det(D\psi)\det(D\Phi_|)}{\det(D\varphi_|)}.$$ From here, I kinda see the denominator of your first definition; I'm not quite sure about the numerator though.

• The OP talked about orthogonal or orthonormal charts, probably you need some more condition to show that this determinant doesn't depend on the choice of the charts. – Integral Nov 8 '16 at 11:34
• Well determinant doesn't depend on a basis, but it is weird that an orthogonal basis are mentioned in the second definition... – Wrath CC Nov 8 '16 at 14:45
• The determinant depend on the charts choice. Here is a brief example: Consider $M=(0,1), N=(0,2)$ and the map $F:(0,1)\to(0,2)$, $F(p)=2p$. Now consider the charts $\varphi:(0,2)\to (0,1)$, $\varphi(t) = t/2$ and $\psi:(0,2^n)\to(0,2)$, $\psi(r) = r/2^{n-1}$. In this case we have $\Phi(t) = \psi^{-1}\circ F\circ \varphi(t) = 2^{n-1}t$. Therefore $\det (D\Phi(t)) = 2^{n-1}$. Clearly I can't write use these charts do define the determinant of $DF$ in an invariant way because I can have different values for each $n$. – Integral Nov 8 '16 at 16:26
• The problem is that my charts are stretching or shortening of some intervals. Orthogonal charts prevent me of doing this. The second definition says precisely that the derivative will be some rotation, and rotations preserve volumes. – Integral Nov 8 '16 at 16:26