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I need to understand what is meant by "tangent space at identity of a Lie group is canonically isomorphic to its Lie algebra" to understand the definition of adjoint representation.

Could any one tell me in detail what is going on in that definition? Thank you.

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This is sometimes used as the definition of the Lie algebra. What other definition are you using? – Qiaochu Yuan Oct 4 '12 at 6:50
It depends on how you define the Lie algebra of a Lie group. Defining it as the tangent space at the identity, equipped with an appropriately defined bracket, is one option. – Marc van Leeuwen Oct 4 '12 at 6:50
up vote 3 down vote accepted

Let $G$ be a matrix Lie group like say $\textrm{GL}_n(\Bbb{C})$ or $\textrm{SL}_n(\Bbb{C})$. An alternative way to define the (real) Lie algebra $\mathfrak{g}$ is to say that it is the set of all $n\times n$ complex matrices $X$ such that

$$e^{tX} \in G \hspace{4mm} \textrm{for all $t \in \Bbb{R}$}.$$

Now using the Lie product formula and some other stuff one can check that the set of all such matrices is in fact a linear space, and we make it into a Lie algebra by defining the commutator of two elements $X,Y$ as $[X,Y] = XY - YX$ where by $XY$ I mean the product of two matrices. This is the reason why I chose $G$ to be a matrix Lie group in the first place so that the bracket is something concrete.

Now I am not so familiar with differential geometry but I will try to explain why $\mathfrak{g}$ now is the tangent space at the identity. Now since everything is happening inside of $\Bbb{C}^{n^2}$ we can think of the tangent space at the identity concretely as the set of all $n \times n$ complex matrices $X$ ( vectors in $\Bbb{C}^{n^2}$ ) such that there is a smooth curve $\gamma(t) \in G$ with $\gamma (0) = I$ and $\gamma'(0) = X$.

We can now prove that given the definition of $\mathfrak{g}$ in terms of the exponential, it is indeed equivalent to being the tangent space at the identity. On one hand if $X \in \mathfrak{g}$ then we can simply take the smooth curve

$$\gamma(t) = e^{tX}$$

that is in $G$ for all $t$ and is such that $\gamma(0) = I$, $\gamma'(0) = X$. For the other direction, suppose that there is a smooth curve $\rho(t)$ in $G$ such that $\rho(0) = I$ and $\rho'(0) = Y$, where $Y$ is some matrix. We wish to show that $Y\in \mathfrak{g}$. Now because the exponential map is a local diffeomorphism about $0 \in\mathfrak{g}$ and $I \in G$,, for $t$ sufficiently small we have that

$$\log( \rho(t)) = \rho(t) - I - \frac{(\rho(t) - I)^2}{2} + \frac{(\rho(t) - I)^3}{3} + \ldots $$

is in $\mathfrak{g}$. Now absolute convergence in the Hilber-Schmidt norm allows us to differentiate log term by term and $\mathfrak{g}$ being a linear space means that the derivative at $t = 0$ is in $\mathfrak{g}$. However the derivative at $t = 0$ is nothing more than the matrix $Y$, so consequently $Y \in \mathfrak{g}$.

This completes the proof that the original definition of the Lie algebra that I gave you is equivalent to it being the tangent space at the identity.

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fabulous.......... – bang Oct 5 '12 at 15:42

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