Doubt in understanding Differentials

Question

I am studying smooth manifolds, and recently I got to know about tangent spaces and the differentials.

Suppose $f:M \to N$ is a smooth map, then it induces a differential between the tangent spaces $df:T_cM \to T_{f(c)}N$ which is a linear transformation. This is what I read.

Now I was trying some examples, and I am stuck here. If I consider $f:\mathbb{R} \to \mathbb{R}$ defined by $f(c)=3c^3$, then the differential map at $x=c$ is $f'(c)=9c^2$. Hence the constant map $9c^2:\mathbb{R} \to \mathbb{R}$ is a linear map, which is not the case. I cannot figure out where I am going wrong.

Answer 1

This is a good question, and an easy misconception!

The differential maps $T_p \mathbb{R} \to T_{f(p)} \mathbb{R}$ for each fixed point $p$. But the tangent space of $\mathbb{R}$ at any point is just $\mathbb{R}$, and this makes it easy to get confused by what is what. It's also confusing that differentials are functions of functions! Remember

$$df_{-} : \mathbb{R} \to \big ( T \mathbb{R} \to T \mathbb{R} \big )$$

That is, for every point $p \in \mathbb{R}$, we have a (distinct!) function $df_p : T_p \mathbb{R} \to T_{f(p)} \mathbb{R}$. This outer function is allowed to be any smooth function (in particular, it can be highly nonlinear!) it is only the inner function that must be linear.

In particular, say $f(p) = 3p^3$. Then you're exactly right, $df = 9p^2 dx$. But what does this mean?

It means for any individual point $p$ we get a map $df_p : T_p \mathbb{R} \to T_{3p^3} \mathbb{R}$. And what is that map?

$$df_p(v) = (9p^2) v$$

This is just multiplication by a scalar (which is linear)! What's confusing is that the choice of scalar depends (nonlinearly) on $p$.

So, as an example:

• $df_1(v) = 9v$
• $df_2(v) = 36v$
• etc.

In general, if you have a smooth function $f : \mathbb{R} \to \mathbb{R}$, then $df_p$ is the linear function which scales by $f'(p)$ (which is just a number).

In even more generality, if you have a smooth function $f : \mathbb{R}^n \to \mathbb{R}^m$, then you may remember we have a jacobian matrix $J$ which has functions as its entries.

Then $df = J$ is a matrix of functions, but when we fix a point $p$ we get $df_p = \left . J \right |_p$ is a matrix with regular old numerical entries. And this $\left . J \right |_p$ is a linear map from $T_p \mathbb{R}^n \to T_{f(p)} \mathbb{R}^m$ (of course, this happens to be the same thing as $\mathbb{R}^n \to \mathbb{R}^m$, but that isn't true for arbitrary manifolds, so it's useful to keep the distinction between $\mathbb{R}^n$ and $T_p \mathbb{R}^n$ in your mind, even though they happen to be the same in this simple case).

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Edit:

Let's take a highly nonlinear function like $\sin(x) : \mathbb{R} \to \mathbb{R}$. Afterwards let's take a nonlinear function from $\mathbb{R}^2 \to \mathbb{R}$ so that we can see a matrix as well.

Then $d\sin(x)_p = \cos(p)dx$. So for any fixed point $p$, say $p = \pi$, we get a linear map

$$d\sin(x)_\pi = v \mapsto \cos(\pi) v$$

that is

$$d\sin(x)_\pi = v \mapsto v$$

which is linear.

Indeed, for any point $p$ you'll get a linear map which comes from scaling $v$ by $\cos(p)$ (which, for fixed $p$, is just a number).

So $$d\sin(x)_1 \approx v \mapsto 0.54 v$$

(which is linear).

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What about in higher dimensions? Let's look at

$$f(x,y) = x^2y$$

Then $df_{p} = df_{(x,y)}$ is the jacobian:

$$ df_{(x,y)} = \left [ \frac{\partial}{\partial x} f \quad \frac{\partial}{\partial y} f \right ] = \left [ 2xy \quad x^2 \right ] $$

Notice the entries of this matrix are nonlinear in the choice of point $p = (x,y)$. However, once we fix a point, say $p = (x,y) = (2,3)$:

$$ df_{(2,3)} = [12 \quad 4] $$

which is a linear map from $T_{(2,3)}\mathbb{R}^2 \to T_{f(2,3)}\mathbb{R}$.

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I hope this helps ^_^

Answer 2

There are two possible misconceptions here. The first is that the map defined as $T_pf:T_pM\to T_{f(p)}N$, sometimes called differential, in reality is a different map than the differential $df$ of a function (this is way I had used two different notations). The first map is also called the tangential map of $f$ at $p$, and it is a linear map that maps vectors from a tangent space to other tangent space.

However the differential is defined by $d_pf v:=\pi_2\circ (T_pf(p,v))$ where $\pi_2$ is the projection of the second argument and $(p,v)\in T_pM$, that is, we have that $T_pf(p,v)=(f(p),d_pf v)$. The maps are very close one to each other but they aren't exactly the same, this is way they generally are both called differential.

Now: the differential at $p$ is a function, if $f:\mathbb{R}^n\to \mathbb{R}^m$ then $d_pf$ is a linear map that can be represented by a Jacobian matrix (the matrix that represents the derivative of $f$ at $p$). If $n=m=1$ then the derivative is just a real number and so the function is just multiplication by an scalar, that is, $d_pf$ will be the map $x\mapsto f'(p)x$.