0
$\begingroup$

I'm trying to learn about smooth manifolds and differential geometry in order to learn the geometric view of classical mechanics. I'm confused about what kind of "object" the differential of a smooth map is.

Set up: Let $M$ be some smooth n-dimensional manifold and let $f: M \to M $ be a smooth map from $M$ onto itself such that a point ${x}\in M$ is mapped to ${y}=f({x})\in M$. Let $(x^1,\dots ,x^n)\in \mathbb{R}^n$ and $(y^1,\dots ,y^n)\in \mathbb{R}^n$ be the coordinate representations of the points ${x}$ and ${y}=f({x})$ in some smooth chart. The differential of $f$ (evaluated at ${x}$) is then a linear map between tangent spaces, $\mathbf{d} f_x: T_x M \to T_{f(x)} M \equiv T_y M$.

my question: Is it correct to say that the matrix representation of $\mathbf{d} f_x$ is given by the Jacobian $\tfrac{\partial y^i}{\partial x^j}$? If so, is $\mathbf{d} f$ a (1,1)-tensor field and $\tfrac{\partial y^i}{\partial x^j}$ are the components of $\mathbf{d} f_x \in T_y M \otimes T^*_x M$? Can we then write it using coordinate basis vector fields $\mathbf{e}_i$, and 1-forms $\boldsymbol{\epsilon}^i$ as:

$$ \mathbf{d} f _x \;=\; \frac{\partial y^i}{\partial x^j} \mathbf{e}_i|_y \otimes \boldsymbol{\epsilon}^j|_x \;=\; \frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial y^i} \otimes \mathbf{d} x^j \quad \in T_y M \otimes T^*_x M \qquad (1) $$

The above is indeed a linear map $T_x M \to T_{f(x)}M$. But $\frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial y^i} = \frac{\partial}{\partial x^j}=\mathbf{e}_j|_x$ so, if the above is correct (is it?), then it is the same thing as

$$ \mathbf{d} f _x \;=\; \frac{\partial}{\partial x^j} \otimes \mathbf{d} x^j \;=\; (\mathbf{e}_j\otimes \boldsymbol{\epsilon}^j)|_x \qquad\qquad (2) $$

And isn't this just the identity tensor field evaluated at the point $x$? What's going on here? Is it possible to expand the differential, either $\mathbf{d}f$ or $\mathbf{d}f_x$, as a linear combination of basis vectors (or rather, tensor products of them)? Or is my attempt to associate $\mathbf{d} f$ with a tensor misguided?

Another question: If $f$ is the map and $y^i$ are just the labels we give to its output, should the jacobian $\frac{\partial y^i}{\partial x^j}$ actually be written as $\frac{\partial f^i}{\partial x^j}$? This is perhaps a small detail with little consequence but I'm just curious. (I'm abusing notation here by using $f$ to denote both the smooth map on $M$ as well as its coordinate representation, $\varphi\circ f \circ \varphi^{-1}$, on $\mathbb{R}^n$)


Edit: question 2: Based on J. Murray's answer, I see why $\mathbf{d} f$ is not really a tensor field since $\mathbf{d} f_x$ maps between tangent spaces of two different points.

Now let the map just be the identity, $f=\text{Id}_{M}$, such that $x=f(x)$ and $\mathbf{d} f_x: T_x M \to T_x M$. And now let $\pmb{x}=(x^1,\dots ,x^n)= \varphi(x)\in\mathbb{R}^n$ and $\pmb{y} =(y^1,\dots ,y^n)=\vartheta(x)\in\mathbb{R}^n$ be coordinate representations of the same point, $x=f(x)\in M_i\subseteq M$, in two different charts, $(M_i,\varphi)$ and $(M_i,\vartheta)$. The coordinates are then related by $\pmb{y} = \widehat{f}(\pmb{x})$ where $\widehat{f} = \vartheta \circ f \circ \varphi^{-1} = \vartheta \circ \varphi^{-1}$ (since $f=\text{Id}$). That is, $\widehat{f}$ is just a transition function. But now using equation (1) with $y=f(x)=x$, we have $\mathbf{d} f_x$ as

$$ \mathbf{d} f _x \;=\; \frac{\partial y^i}{\partial x^j} (\mathbf{e}_i \otimes \boldsymbol{\epsilon}^j)|_x \;=\; \frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial x ^i} \otimes \mathbf{d} x^j \;=\; \frac{\partial y^i}{\partial x^j} \frac{\partial}{\partial y ^i} \otimes \mathbf{d} y^j \quad \in T_x M \otimes T^*_x M $$ The above, if it is correct, is a (1,1)-tensor on the vector space $T_x M$, expressed using two different coordinate basis vectors and 1-forms; one basis associated with coordinates $x^i$ and another associated with the coordinates $y^i$ (both coordinates for the same point). The above is also equivalent to

$$ \mathbf{d}( \text{Id})_x \;\equiv \;\mathbf{d} f _x \;=\; \frac{\partial}{\partial x ^i} \otimes \mathbf{d} y^i $$

where $\mathbf{d} f_x$ is really $\mathbf{d} (\text{Id})_x$. So when we have some coordinate transformation on $\mathbb{R}^n$, $\pmb{y}=\widehat{f}(\pmb{x})$, and we write the Jacobian matrix, $\frac{\partial y^i}{\partial x^j}$ or $\frac{\partial \pmb{y}}{\partial \pmb{x}}$, is this matrix actually a coordinate representation of the differential of the identity map on $M$, but expressed using a mix of two different charts?

$\endgroup$

1 Answer 1

1
$\begingroup$

Is it correct to say that the matrix representation of $\mathbf{d} f_x$ is given by the Jacobian $\tfrac{\partial y^i}{\partial x^j}$?

Yes, that's correct.

If so, is $\mathbf{d} f$ a (1,1)-tensor field and $\tfrac{\partial y^i}{\partial x^j}$ are the components of $\mathbf{d} f_x \in T_y M \otimes T^*_x M$? [...] Or is my attempt to associate $\mathbf{d} f$ with a tensor misguided?

Yes, it is misguided. A tensor field $S$ on $M$ can be understood as an assignment of a tensor to each point of $M$. In particular, it is meaningful to talk about the "element" of $S$ at a point $p\in M$, which we might denote $S_p$. Contrast that with $\mathrm df$, which (if we interpret it as eating two vectors and spitting out a scalar) would eat a vector at $p$ and a vector at a different point $f(p)$. In that sense, we cannot speak of the "element" of $\mathrm df$ at any particular point.

Put differently, a $(r,s)$-tensor field can be understood as a map which eats $r$ covectors and $s$ vector fields and spits out a scalar. Your object eats (co)vectors living at different points, not (co)vector fields.

If $f$ is the map and $y^i$ are just the labels we give to its output, should the jacobian $\frac{\partial y^i}{\partial x^j}$ actually be written as $\frac{\partial f^i}{\partial x^j}$?

The right way to express the Jacobian is $\partial_j\big(y^i\circ f \circ x^{-1}\big)$, where $x$ and $y$ are the relevant coordinate chart maps, $y\circ f \circ x^{-1}$ is a map from $\mathbb R^n \rightarrow \mathbb R^n$, and $\partial_j$ denotes differentiation with respect to the $j^{th}$ slot. By convention, we typically define the notation $\partial_j (g\circ x^{-1}) \equiv \partial g/\partial x^j$, and so we might apply this to yield the expression $\partial(y^i \circ f)/\partial x^j$. The notation is then further condensed by (as you say) identifying $y^i\circ f$ simply as $y^i$ to yield the standard expression for the Jacobian.

$\endgroup$
4
  • $\begingroup$ I see. So its wrong to use the word tenor (field) for $\mathbf{d} f_x$ or $\mathbf{d} f$. But if we just take the differential evaluated at a point, $\mathbf{d} f_x$, is it accurate to express this as I did in my first equation? This tensor-looking-thing indeed eats a vector in $T_x M$ and spits out a vector in $T_{f(x)}M$, as you said. Im just trying to figure out if there's a way to write the object $\mathbf{d} f_x$ in terms of the Jacobian $\frac{\partial y^i}{\partial x^j}$, in the way that we write a vector, $\mathbf{u}$, using its coordinates as $\mathbf{u}=u^i \mathbf{e}_i$ $\endgroup$
    – J Peterson
    Oct 4, 2022 at 20:02
  • 1
    $\begingroup$ @JPeterson Sure, if you'd like you can think of $\mathrm df_p$ as an element of $V_{f(p)} \otimes V^*_p$. Just note that such an object is not a tensor (field). $\endgroup$
    – J. Murray
    Oct 4, 2022 at 20:06
  • $\begingroup$ your answer made me think of another related question that I added to my post. You are already answered my original question but perhaps you have some thoughts. $\endgroup$
    – J Peterson
    Oct 4, 2022 at 22:17
  • $\begingroup$ @JPeterson I would agree with your assessment - after all, given overlapping charts $(U,x)$ and $(V,y)$, the chart transition map $y\circ x^{-1}:x(U\cap V) \rightarrow y(U\cap V)$ can also be written as $y \circ \mathrm{id}_M \circ x^{-1}$. $\endgroup$
    – J. Murray
    Oct 5, 2022 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.