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Let $G$ and $H$ be two Lie groups and $\rho: G \to H$ be a homomorphism. How to differentiate $\rho$ to obtain a Lie algebra homomorphism $d\rho_e: T_eG \to T_eH$?

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3 Answers 3

up vote 6 down vote accepted

There are a number of equivalent ways to define the derivative of a map of manifolds, since there are a number of equivalent ways to define the tangent space of a manifold at a point.

One way which makes the definition of derivative particularly simple is as follows: Let $p$ be a point on a manifold $M$, and let $C^\infty(M)$ be the algebra of smooth real-valued functions on $M$. The tangent space $T_pM$ can be thought of as the space of derivations at $p$; that is, $T_pM$ is the space of $\mathbb{R}$-linear maps $\delta: C^\infty(M)\to\mathbb{R}$ which satisfy the rule $\delta(fg) = \delta(f)g(p) + f(p)\delta(g)$. Given a local coordinate system $x_1,\ldots,x_m$ near $p$, a basis for $T_pM$ is given by the operators $\frac{\partial}{\partial x_i}|_p$.

It should be noted that with this definition, a vector field (i.e. a smooth choice of tangent vector for each point) is then the same thing as a derivation of $C^\infty(M)$, i.e. an $\mathbb{R}$-linear map $\Delta:C^\infty(M)\to C^\infty(M)$ satisfying the Leibniz rule $\Delta(fg) = \Delta(f)g + f\Delta(g)$. Given a derivation $\Delta$, the tangent vector $\delta\in T_p M$ is the map $C^\infty(M)\to\mathbb{R}$ given by composing $\Delta$ with evaluation at $p$.

With this description of the tangent space, the derivative $d\varphi_p:T_p M\to T_{\varphi(p)}N$ of a map $\varphi:M\to N$ at a point $p\in M$ can be described as follows: first, $\varphi$ induces a map of algebras $\varphi^*:C^\infty(N)\to C^\infty(M)$ by $\varphi^*(g)=g\circ\varphi$. The derivative is then given by $d\varphi_p(\delta) = \delta\circ\varphi^*$.

For the purposes of computation, the derivative can be described in local coordinates as follows: suppose $x = (x_1,\ldots,x_m)$ is a local coordinate system for $M$ at $p$, and $y = (y_1,\ldots,y_n)$ is a local coordinate system for $N$ at $\varphi(p)$. Then $\varphi$ is given near $p$ in terms of these coordinates by functions $\varphi_1(x),\ldots,\varphi_n(x)$ -- that is, $\varphi_j$ is the $y_j$-component of $\varphi$. The differential $d\varphi_p$ is then defined on the basis $\frac{\partial}{\partial x_i}|_p$ by $\displaystyle d\varphi_p\left(\frac{\partial}{\partial x_i}|_p\right) = \sum_{j=1}^n \frac{\partial\varphi_j}{\partial x_i}|_p \frac{\partial}{\partial y_j}|_{f(p)}$.

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Your definition of T_p(M) doesn't seem to refer to p. – Qiaochu Yuan Jan 2 '11 at 18:11
Hi Brad, thank you very much. – user Jan 2 '11 at 18:15
It does refer to $p$, as the "Leibniz rule at $p$" involves evaluation at $p$. – Brad Jan 2 '11 at 19:36
Should $f(p)$ in the last paragraph be $\varphi(p)$? If we have a map $\varphi: M \to N$, $x = (x_1,\ldots,x_m)$ is a local coordinate system for $M$ at $p$, and $y = (y_1,\ldots,y_n)$ is a local coordinate system for $N$ at $\varphi(p)$, what is the explicite form of the induced map $\varphi^*:C^\infty(N)\to C^\infty(M)$ which is defined by $\varphi^*(g)=g\circ\varphi$ using local coordinates? Why $\displaystyle d\varphi_p\left(\frac{\partial}{\partial x_i}|_p\right) = \sum_{j=1}^n \frac{\partial\varphi_j}{\partial x_i}|_p \frac{\partial}{\partial y_j}|_{f(p)}$? Thank you very much. – user Jan 4 '11 at 17:38
Hi Qiaochu, thank you very much. – user Jan 4 '11 at 17:40

Given any two manifolds and a map between them, there is automatically induced a map on the tangent spaces at corresponding points. The only new thing here is to check that the induced map is actually a homomorphism of Lie algebras.

Let $v_1, v_2 \in T_e(G)$. We can extend them to left-invariant vector fields $V_1, V_2$ on $G$. In fact, we identify $T_e(G)$ with the space of left-invariant vector fields on $G$. The bracket is just given by the usual Lie bracket $[V_1, V_2]$. (One has to check that the Lie bracket preserves left-invariance; this is a general fact about it. In detail, if $X, Y$ are manifolds and $f: X \to Y$ a map, then vector fields $V$ on $X$ and $W$ on $Y$ are said to be $f$-related if $f_*(X) = Y$ wherever both sides are defined. It is a basic fact that if $X_1, Y_1$ and $X_2, Y_2$ are $f$-related, then $[X_1, X_2]$ and $[Y_1, Y_2]$ are $f$-related.)

Suppose $v_1, v_2$ get sent to $w_1, w_2 \in T_e(H)$. Extend $w_1, w_2$ to invariant vector fields $W_1, W_2$ on $H$. Then $V_1, V_2$ are $f$-related to $W_1, W_2$ by left-invariance. It follows that $[V_1, V_2]$ is $f$-related to $[W_1, W_2]$. This implies that the values at $e$ match up. This is precisely what you are asking for.

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My guess is that OP is actually asking a simpler question, that is what's the definition of the kind of derivative used here. – Noah Snyder Jan 2 '11 at 15:40
Thank you very much. – user Jan 2 '11 at 16:11
Hi Akhil, thank you very much. But I do not understand the induced map on tangent spaces. Given any two manifolds and a map between them, what is the induced map on the tangent spaces at corresponding points? – user Jan 2 '11 at 16:59
Hi Noah, thank you very much. – user Jan 2 '11 at 17:07

Regarding the comments to Akhil's answer:

If $x$ is a point in a manifold $M$, the tangent space $T_x M$ can be defined in various ways, one of which is to think of a tangent vector as an equivalence class of curves through $x$ (two curves being equivalent if they pass through the point with the same velocity). If $\gamma$ is such a curve (corresponding to a tangent vector to $M$ at $x$, call it $v$), then a function $F:M \to N$ takes $\gamma$ to a curve $F\circ \gamma$ in $N$, which passes through the point $F(x)$ with a certain velocity, corresponding to a tangent vector to $N$ at $F(x)$, and it is this tangent vector that is denoted by $dF(v)$. The mapping from $v \in T_x M$ to $dF(v) \in T_{F(x)} N$ is the induced mapping on the tangent spaces.

In local coordinates we can write $F(x) = (F_1(x_1,\dots,x_m), \dots, F_n(x_1,\dots,x_m))$, and the linear transformation $dF$ simply becomes the $n \times m$ matrix of derivatives, $(\partial F_i/\partial x_j)$.

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Hi Hans, thank you very much. Now I understand better about this question. – user Jan 2 '11 at 18:14

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