# The Unapologetic Mathematician

## The Jacobian

Now that we’ve used exterior algebras to come to terms with parallelepipeds and their transformations, let’s come back to apply these ideas to the calculus.

We’ll focus on a differentiable function $f:X\rightarrow\mathbb{R}^n$, where $X$ is itself some open region in $\mathbb{R}^n$. That is, if we pick a basis $\{e_i\}_{i=1}^n$ and coordinates of $\mathbb{R}^n$, then the function $f$ is a vector-valued function of $n$ real variables $x^1,\dots,x^n$ with components $f^1,\dots,f^n$. The differential, then, is itself a vector-valued function whose components are the differentials of the component functions: $df=df^ie_i$. We can write these differentials out in terms of partial derivatives:

$\displaystyle df^i(x^1,\dots,x^n;t^1,\dots,t^n)=\frac{\partial f^i}{\partial x^1}\bigg\vert_{(x^1,\dots,x^n)}t^1+\dots+\frac{\partial f^i}{\partial x^n}\bigg\vert_{(x^1,\dots,x^n)}t^n$

Just like we said when discussing the chain rule, the differential at the point $(x^1,\dots,x^n)$ defines a linear transformation from the $n$-dimensional space of displacement vectors at $(x^1,\dots,x^n)$ to the $n$-dimensional space of displacement vectors at $f(x^1,\dots,x^n)$, and the matrix entries with respect to the given basis are given by the partial derivatives.

It is this transformation that we will refer to as the Jacobian, or the Jacobian transformation. Alternately, sometimes the representing matrix is referred to as the Jacobian, or the Jacobian matrix. Since this matrix is square, we can calculate its determinant, which is also referred to as the Jacobian, or the Jacobian determinant. I’ll try to be clear which I mean, but often the specific referent of “Jacobian” must be sussed out from context.

So, in light of our recent discussion, what does the Jacobian determinant mean? Well, imagine starting with a $n$-dimensional parallelepiped at the point $(x^1,\dots,x^n)$, with one side in each of the basis directions, and positively oriented. That is, it consists of the points $(x^1+t^1,\dots,x^n+t^n)$ with $t^i$ in the interval $[0,\Delta x^i]$ for some fixed $\Delta x^i$. We’ll assume for the moment that this whole region lands within the region $X$. It should be clear that this parallelepiped is represented by the wedge

$\displaystyle(\Delta x^1e_1)\wedge\dots\wedge(\Delta x^ne_n)=(\Delta x^1\dots\Delta x^n)e_1\wedge\dots\wedge e_n$

which clearly has volume given by the product of all the $\Delta x^i$.

Now the function $f$ sends this cube to a sort of curvy parallelepiped, consisting of the points $f(x^1+t^1,\dots,x^n+t^n)$, with each $t^i$ in the interval $[0,\Delta x^i]$, and this image will have some volume. Unfortunately, we have no idea as yet how to measure such a volume. But we might be able to approximate it. Instead of using the actual curvy parallelepiped, we’ll build a new one. And if the $\Delta x^i$ are small enough, it will be more or less the same set of points, with the same volume. Or at least close enough for our purposes. We’ll replace the curved path defined by

$\displaystyle f(x^1,\dots,x^i+t,\dots,x^n)\qquad0\leq t\leq\Delta x^i$

by the displacement vector between the two endpoints:

$\displaystyle f(x^1,\dots,x^i+\Delta x^i,\dots,x^n)-f(x^1,\dots,x^i,\dots,x^n)$

and use these new vectors to build a new parallelepiped

$\displaystyle\left(f(x^1+\Delta x^1,\dots,x^n)-f(x^1,\dots,x^n)\right)\wedge\dots\wedge\left(f(x^1,\dots,x^n+\Delta x^n)-f(x^1,\dots,x^n)\right)$

But this is still an awkward volume to work with. However, we can use the differential to approximate each of these differences

$\displaystyle\begin{aligned}f(x^1,\dots,x^k+\Delta x^k,\dots,x^n)&-f(x^1,\dots,x^k,\dots,x^n)\\&\approx df(x^1,\dots,x^n;0,\dots,\Delta x^k,\dots,0)\\&=\Delta x^kdf(x^1,\dots,x^n;0,\dots,1,\dots,0)\\&=\Delta x^kdf^i(x^1,\dots,x^n;0,\dots,1,\dots,0)e_i\\&=\Delta x^k\frac{\partial f^i}{\partial x^k}\bigg\vert_{(x^1,\dots,x^n)}e_i\end{aligned}$

with no summation here on the index $k$.

Now we can easily calculate the volume of this parallelepiped, represented by the wedge

$\displaystyle\left(\Delta x^1\frac{\partial f^i}{\partial x^1}\bigg\vert_{(x^1,\dots,x^n)}e_i\right)\wedge\dots\wedge\left(\Delta x^n\frac{\partial f^i}{\partial x^n}\bigg\vert_{(x^1,\dots,x^n)}e_i\right)$

which can be rewritten as

$\displaystyle\left(\Delta x^1\dots\Delta x^n\right)\left(\frac{\partial f^i}{\partial x^1}\bigg\vert_{(x^1,\dots,x^n)}e_i\right)\wedge\dots\wedge\left(\frac{\partial f^i}{\partial x^n}\bigg\vert_{(x^1,\dots,x^n)}e_i\right)$

which clearly has a volume of $\left(\Delta x^1\dots\Delta x^n\right)$ — the volume of the original parallelepiped — times the Jacobian determinant. That is, the Jacobian determinant at $(x^1,\dots,x^n)$ estimates the factor by which the function $f$ expands small volumes near that point. Or it tells us that locally $f$ reverses the orientation of small regions near the point if the Jacobian determinant is negative.

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Posted by John Armstrong | Analysis, Calculus

## 22 Comments	»

1. [...] « Previous | [...]

Pingback by | November 12, 2009 | Reply

2. [...] Posts The Jacobian of a CompositionThe JacobianThe Cross Product and PseudovectorsThe Hodge StarSunday Samples 146An Example of a [...]

Pingback by | November 13, 2009 | Reply

3. [...] this case, the Jacobian transformation is just the function itself, and so the Jacobian determinant is nonzero if and only if the matrix [...]

Pingback by | November 17, 2009 | Reply

4. [...] the theorem that I promised. Let be continuously differentiable on an open region , and . If the Jacobian determinant at some point , then there is a uniquely determined function and two open sets and so [...]

Pingback by | November 18, 2009 | Reply

5. [...] rank if and only if the leading matrix is invertible. And this is generalized to asking that some Jacobian determinant of our system of functions is [...]

Pingback by | November 19, 2009 | Reply

6. [...] That is, the new component functions are just the coordinate functions . We can easily calculate the Jacobian matrix [...]

Pingback by | November 20, 2009 | Reply

7. [...] cases, we know that the inverse function exists because of the inverse function theorem. Here the Jacobian determinant is simply the derivative , which we’re assuming is everywhere [...]

Pingback by | January 5, 2010 | Reply

8. [...] differentiable function defined on an open region . Further, assume that is injective and that the Jacobian determinant is everywhere nonzero on . The inverse function theorem tells us that we can define a continuously [...]

Pingback by | January 6, 2010 | Reply

9. [...] Geometric Interpretation of the Jacobian Determinant We first defined the Jacobian determinant as measuring the factor by which a transformation scales infinitesimal pieces of -dimensional [...]

Pingback by | January 8, 2010 | Reply

10. [...] again, in the sense that if one of these integrals exists then so does the other, and their values are equal. The function plays the role of the absolute value of the Jacobian determinant. [...]

Pingback by | August 2, 2010 | Reply

11. [...] and take the th partial derivative of that function. And this is precisely the definition of the Jacobian of this transition [...]

Pingback by | April 1, 2011 | Reply

12. [...] from multivariable calculus of a linear map that takes tangent vectors to tangent vectors: the Jacobian, which we saw as a certain extension of the notion of the derivative. We will find that our map is [...]

Pingback by | April 6, 2011 | Reply

13. [...] dimensions of and , respectively. Like we saw for coordinate transforms in place, this is just the Jacobian [...]

Pingback by | April 6, 2011 | Reply

14. [...] are the components of the Jacobian matrix of the transition function . What does this mean? Well, consider the linear [...]

Pingback by | April 13, 2011 | Reply

15. [...] function theorem from multivariable calculus: if is a map defined on an open region , and if the Jacobian of has maximal rank at a point then there is some neighborhood of so that the restriction is [...]

Pingback by | April 14, 2011 | Reply

16. [...] , and the Jacobian of [...]

Pingback by | April 15, 2011 | Reply

17. [...] is, the two forms differ at each point by a factor of the Jacobian determinant at that point. This is the differential topology version of the change of basis formula for top [...]

Pingback by | July 8, 2011 | Reply

18. [...] look at what happens when and is a singular -cube. Since it has a Jacobian at each point in the unit cube, and we’ll keep things simple by assuming that it’s [...]

Pingback by | August 3, 2011 | Reply

19. [...] little thought gives us our answer: is the Jacobian determinant of the coordinate transformation from one patch to the other. Indeed, we use the Jacobian to change [...]

Pingback by | August 29, 2011 | Reply

20. [...] the (local) orientation form on differs from the (local) orientation form on by a factor of the Jacobian determinant of the function with respect to these coordinate maps. This repeats what we saw in the case of [...]

Pingback by | September 1, 2011 | Reply

21. [...] manifold to another. Since is both smooth and has a smooth inverse, we must find that the Jacobian is always invertible; the inverse of at is at . And so — assuming is connected — [...]

Pingback by | September 12, 2011 | Reply

22. [...] So let’s see how this form changes; if is another coordinate patch, we can assume that by restricting each patch to their common intersection. We’ve already determined that the forms differ by a factor of the Jacobian determinant: [...]

Pingback by | October 6, 2011 | Reply

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