Background: Let $f: \mathbb R^n \to \mathbb R^m$ be a differentiable vector-valued function, with components $f_i: \mathbb R^n \to \mathbb R, \textrm{ for }1\le i\le m$. Then the Jacobian matrix is the $m \times n$ matrix $$\textbf{J} = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \cdots & \dfrac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial f_m}{\partial x_1} & \cdots & \dfrac{\partial f_m}{\partial x_n} \end{bmatrix}$$

As is well-known, in the case in which $n = m$ the Jacobian matrix is square, and its determinant, the Jacobian determinant, is the scalar $J = \det \textbf{J}$. The Jacobian determinant has many uses, including when using a substitution during integration.

But what if $m \ne n$? In this case, $\textbf{J}$ is not square, so $\det \textbf{J}$ is not defined. However, in this case, the quantity $\left(\det \textbf{J} \textbf{J}^\top \right)^{1/2}$ is still meaningful, and generalizes the idea (and some of the uses) of the Jacobian determinant:

  1. For example, consider a function $\vec r: \mathbb R^2 \to \mathbb R^3$, which we can regard as a parametrization of a surface in $\mathbb R^3$. If we write $\vec r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$, then the Jacobian matrix is $\textbf{J} = \begin{bmatrix} x_u & y_u & z_u \\ x_v & y_v & z_v \end{bmatrix}$; in this situation it can be shown that $\left(\det \textbf{J} \textbf{J}^\top \right)^{1/2} = \| \vec r_u \times \vec r_v \|$. We recognize the quantity that relates area in the surface to area in the domain: $dS = \| \vec r_u \times \vec r_v \| \, dA$.
  2. On the other hand consider a function $\vec r: \mathbb R^1 \to \mathbb R^n$, which we can interpret as a parametrization of a path in $\mathbb R^n$. If we write $\vec r(t) = \langle x_1(t), x_2(t), \dots, x_n(t) \rangle$ then in this case $\left(\det \textbf{J} \textbf{J}^\top \right)^{1/2} = \left| \frac{d\vec r}{dt} \right|$. Once again we recognize the quantity that relates arc length in the path to length in the domain: $ds = \left| \frac{d\vec r}{dt} \right| \, dt$.
  3. Finally we note that if $n = m$ then $\left(\det \textbf{J} \textbf{J}^\top \right)^{1/2}$ is identical to the "usual" Jacobian determinant $J$.

My question:

Does the quantity $\left(\det \textbf{J} \textbf{J}^\top \right)^{1/2}$ have a standard name in the case of a non-square Jacobian matrix $\textbf{J}$? The phrase "Jacobian determinant" seems, as far as I can tell, to only be used in the context of square matrices, despite the fact that the quantity generalizes quite naturally to the non-square case, and encompasses several "special formulas" in a single neat form. But I am unaware of any established terminology for this quantity. Is there one?

  • $\begingroup$ In case anyone's interested, I have apparently been wondering about this (in at least one special case) for almost two years: see math.stackexchange.com/questions/3805486/… $\endgroup$
    – mweiss
    Jul 28, 2022 at 19:56
  • $\begingroup$ By the way, In your first definition of the Jacobian you use the convention that each row $i$ of $\mathbf{J}$ is the transpose of the gradient of the $i$-th coordinate of $f$, $\nabla f_i^T$. But in your first example you have written down the transpose of this, so that each column is the gradient of the coordinates. The latter representation makes current answer relevant. Was this intentional? $\endgroup$ Aug 11, 2022 at 12:00

1 Answer 1


Im not sure if you're already aware of this, but $J^TJ$ is actually a matrix representation of the metric tensor $g$ in some coordinates. Indeed, recall that the column vectors in the Jacobian matrix represent the new basis vectors in terms of the old ones, hence $(J^TJ)_{ij}=e_i.e_j$.

That is, the components of $J^TJ$ exactly represent the dot product of the new basis vectors, which is exactly what the components of the metric tensor represent.

Now when $n\ne m$ like in your case then $J^TJ$ represents the induced metric, and hence its determinant is just the determinant of the induced metric.

  • 1
    $\begingroup$ The question was about $JJ^\top$, though, not $J^\top J$. These are not the same (they are not even the same size matrix) -- unless there are different conventions about which matrix is "the Jacobian"? I can definitely see that as being a possibility. $\endgroup$
    – mweiss
    Jul 28, 2022 at 19:59

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