# jacobian on a line

How does the interpretation of the Jacobian as the stretch factor apply when looking at a function on a subset. Say my function is projection of the plane to the x-axis, $$(x,y)\mapsto (x,0)$$. As a function from the plane to plane it collapses volumes so the Jacobian is 0 like I expect, $$|\begin{pmatrix}1&0\\0&0\end{pmatrix}|=0.$$ But what if this projection is viewed as a function from the line $$y=x$$? It is one-to-one here. It looks like a segment of volume/length $$\sqrt{2}$$ maps to one of volume/length $$1$$, so I would expect a "Jacobian-like" object to be $$1/\sqrt{2}$$. What is the Jacobian like object?

From reading on the Internet I have the impression this has something to do with differential forms which I have not studied, only multivariable calculus, and I already graduated. Is it possible to understand, at least with simple things like lines, without that theory?

• The line $y=x$ can be written as $(x,x)$ in $\mathbb{R}^2$. In particular this a one dimensional subspace, so the jacobian is just a number. Mar 31 at 22:39
• @Chris But you can't use the usual jacobian formula, right? If I try to calculate the Jacobian of $(x,x)\mapsto (x,0)$ I would just get 0, it would not take into account the way you re-wrote the line Mar 31 at 22:41
• To me this issue becomes really clear in the context of smooth manifolds. I think simple cases like this one are honestly the hardest to grasp. I’ll try and find time to write up a detailed answer, but rn the reason your jacobian isn’t giving you what you expect is because the line y=x isn’t and open set of $\mathbb{R}^2$, so you can’t really naively restrict your function to this line Mar 31 at 23:38

The answer is linear algebra, not calculus. You want to write down the (in your example, $$1\times 1$$) square matrix representing the projection map from the domain subspace to the range. In particular, you want to write down an orthonormal basis for each and consider the matrix whose $$j$$th column is the projection of the $$j$$th basis vector for the domain expressed in terms of the basis for the range. You then take the determinant of this matrix.
In your example, the basis for the domain is $$e_1=\frac1{\sqrt2}(1,1)$$, the basis for the range is $$f_1=(1,0)$$, and the projection, as you says, maps $$e_1$$ to $$\frac1{\sqrt2}f_1$$, so the matrix is just $$\begin{bmatrix}\frac1{\sqrt2}\end{bmatrix}$$.
Let the jacobian matrix for the map $$f$$ be $$D_x$$ at the point $$x \in \mathbb R^n$$. Consider any point $$x_0$$. Take any unit vector $$v$$ along any direction. What does the map $$f$$ do to the direction $$v$$ around the point $$x_0$$ ? It rotates and possibly stretches this direction to $$D_{x_{0}} v$$. What happens to the length of the unit vector we considered ? It changes to $$||D_{x_{0}} v||_2$$. This is your jacobian like object. In this case take $$v = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})^{'}$$, a unit vector along the line $$y = x$$. Then $$||D_{x_{0}} v||_2 = \frac{1}{\sqrt{2}}$$, as expected.
You may consider any affine space $$S$$ passing through $$x_0$$. Pick up basis set of unit vectors $$v_1, ..., v_n$$ for the space $$S$$. These vectors form a unit parallelepiped in $$S$$. This parallelepiped gets transformed to a parallelepiped created by $$D_{x_{0}} v_1, ..., D_{x_{0}} v_n$$. Form the matrix $$M$$ having column as these transformed vectors. Then the volume of this parallelepiped in $$S$$ is $$\sqrt{det(M'M)}$$. This gives you the jacobian type object for $$S$$.