I'm reading Allan Pollack's Differential Topology and got stuck on this argument: In the second paragraph of page 9, section 1.2 he said

"Note that if $f:U\to \mathbf{R^m}$ is itself a linear map $L$, then $df_x=L$ for all $x\in U$. In particular, the derivative of the inclusion map of $U$ into $\mathbf{R^n}$ at any point $x\in U$ is the identity transformation of $\mathbf{R^n}$."

I don't quite understand his first sentence. How could $f=L$ and $L$ linear imply $df_x=L$? A counter example is $f(x)=x$ linear, then $f'(x)=1 \neq L$.

Thanks a lot for everyone's help!

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    $\begingroup$ The differential of a one-dimensional function $x\mapsto f(x)$ is the linear map $df_x:v\mapsto f'(x)v$ (well, family of linear maps). Thus, in your case, $f'(x)=1$ implies the differential is $v\mapsto v$, which is in fact the same as $f$, namely the identity map. $\endgroup$ – anon May 15 '13 at 7:59
  • $\begingroup$ Thanks anon, that makes sense. However it sounds a bit like black magic to me. f'(x)=1 implies f(x)=1 and 1 denotes identity map in the later case. Is it some theory that I'm unfamiliar with? $\endgroup$ – Evariste May 15 '13 at 10:24
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    $\begingroup$ $f'(x)=1$ does not imply $f(x)=1$, rather it implies $df_x={\rm Id}$ or equivalently the matrix of $df_x$ is simply $[1]$. The only theory behind it is the theory I have described already - namely, the differential needs to be thought of as a linear map, and there is a difference between a linear map and the matrix representing it. $\endgroup$ – anon May 15 '13 at 11:01
  • $\begingroup$ I see your points here. Thanks a lot! $\endgroup$ – Evariste May 15 '13 at 14:49

Formally, the situation is like this: there is a function $\phi:{\cal M}\to{\cal N}$ (between manifolds), and we want something to describe infinitesimal change. We associate to each $x\in{\cal M}$ a linear map, the differential, which is a transformation of tangent spaces $df_x:T_x{\cal M}\to T_{f(x)}{\cal N}$.

$\hskip 0.6in$ differential

Image source: Wikipedia.

In Euclidean space, points themselves may be viewed as vectors and hence added or subtracted, and so the manifold and all of its tangent spaces may be identified together. That is, every tangent vector exists as a point in the original space (codomain). If $f:{\bf R}^n\to{\bf R}^m$ is differentiable, then the differential is the "directional derivative" as a linear function of the "direction." Explicitly, the matrix of this linear map $df_x$ is given by the Jacobian. The write-up says that the Jacobian matrix of the linear map $L:x\mapsto Ax$ is just $A$, hence $dL_x=L$ is an equality of maps for all $x$.

In the one-dimensional case, $f:{\bf R}\to{\bf R}$, the matrix of a linear map is $1\times1$, so essentially just a scalar value. The scalar is in fact $f'(x)$, so the differential is $df_x:v\mapsto f'(x)v$. In particular, $f(x)=x$ implies $f'(x)=1$ so $df_x:v\mapsto1v$ is the identity map, i.e. the same as $f$.

  • $\begingroup$ Thanks anon! This answer looks good! I still have to take some time to get my head around this. Thanks a lot for the help! $\endgroup$ – Evariste May 15 '13 at 10:27
  • $\begingroup$ Hello anon! I think it makes more sense to me now! I was stuck on why f=L => df_x=L. But it makes sense in terms of Jacobian matrix. Thanks so much for the help! $\endgroup$ – Evariste May 15 '13 at 14:54

On top of the other answer mentioned here, I found it useful to look at the definition of $df_x$. According to Milnor:

$$ df_x(h) = \frac{f(x+th) - f(x)}{t}$$

As $f$ is linear:

$$\begin{align} df_x(h) &= \lim_{t \to 0} \frac{f(x+th) - f(x)}{t} \\ &= \lim_{t \to 0} \frac{f(x) + f(th) - f(x)}{t} \\ &= \lim_{t \to 0} \frac{f(x) + tf(h) - f(x)}{t} \\ &= \lim_{t \to 0} \frac{tf(h)}{t} \\ &= \lim_{t \to 0} f(h) \\ &= f(h) \end{align}$$


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