Given $R^n$ and $R^k$ and $L$ is a linear map from $R^n$ to $R^k$. I was told that the differential of $L$ at $p$, $dL_p:T_pR^n \to T_{L(p)}R^k$ is $L$ itself. Here $dL_p(v)(g) = v(g\circ L)\ \forall v \in T_pR^n$. This is really confusing since $L$ and $dL_p$ does not even have the same source space. In what sense should they be "equal"?
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What is a "linear map" between smooth manifolds? Usually linear maps are defined between linear spaces, where the tangent space at any point can be canonically identified with the space itself. Such maps are automatically differentiable, with the differential coinciding with the map itself. – lisyarus Nov 02 '16 at 16:03
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Sorry that I have edited the question. – Keith Nov 02 '16 at 16:07
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1$\Bbb R^n \simeq T_{p}\Bbb R^n$ – Nov 02 '16 at 16:08
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1No, $L$ itself has $L$ as a differential. Synthetically, this is the simple remark that at every $x$, $$L(x+h)=L(x)+\color{red}{L(h)}+o(h)$$ when $h\to0$. Of course here, the $o(h)$ term is actually zero hence the identity above indicates that, for every $h$, $\color{red}{L(h)}=dL_x(h)$, qed. – Did Nov 02 '16 at 16:12
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The source space of dL_x is the tangent space of $R^k$ at $p$ rather than $R^k$ it self. – Keith Nov 02 '16 at 16:14
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I see your point. @Bye_World – Keith Nov 02 '16 at 16:17
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@lisyarus is this related to why the differential of left multiplication in the general linear group is also left multiplication? – Ekhin Taylor R. Wilson Nov 07 '19 at 10:04
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The point is that if $V$ is any finite-dimensional vector space, there is a canonical (i.e., basis-independent) isomorphism between each tangent space $T_pV$ and $V$ itself. Using the definition of the tangent space $T_pV$ as the space of derivations of $C^\infty(V)$ at $p$, this isomorphism is given by sending $v\in V$ to the derivation $D_v\colon C^\infty(V)\to \mathbb R$, defined by $$ D_v f = \left.\frac{d}{dt}\right|_{t=0} f(p+tv). $$
Once you make the identifications $T_p\mathbb R^n\cong \mathbb R^n$ and $T_{L(p)}\mathbb R^k\cong \mathbb R^k$, the equation $DL_p = L$ falls out quickly from the definition of the differential.
Jack Lee
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Jack Lee, is this related to why the differential of left multiplication in the general linear group is also left multiplication? – Ekhin Taylor R. Wilson Nov 07 '19 at 10:03
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1@EkhinTaylorR.Wilson: Yes. Left multiplication by a fixed matrix is the restriction to an open subset (i.e., $GL(n,\mathbb R)$) of a linear map on the vector space of all $n\times n$ real matrices. – Jack Lee Nov 07 '19 at 19:14
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Thanks! You can post your comment as an answer to my question. – Ekhin Taylor R. Wilson Nov 08 '19 at 05:38