What is meant by "tangent space at identity of a Lie group is canonically isomorphic to its Lie algebra"? 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.
 A: 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.
A: I can do much more than that, here. Let's start at the beginning.
A differential manifold $G$ is a continuum that has been given enough infrastructure to allow for meaningful notions of continuity, contiguity (by its topological structure), and differentiation (by its differential structure); so that one can talk about connected, continuous, differentiable paths. If $t ∈ ℝ ↦ g(t) ∈ G$ is such a path, then by both definition and construction, $g'(t)$ is tangent to the curve at $g(t)$, and $g'(t) ∈ T_{g(t)}G$. That's effectively how tangent spaces are defined and constructed. For each $g ∈ G$, the tangent space $T_gG$ is the set of all tangents to all differentiable curves passing through $g$. So, it's natural to denote the generic element of $T_gG$ as $\dot g$, which I'll do here. Together, the family $TG ≡ \{(g,\dot g): g ∈ G, \dot g ∈ T_gG)\}$ is also called the tangent bundle of $G$.
Similarly, for a function $φ: G → G'$, between differential manifolds $G$ and $G'$, one can define the differential $φ_*: TG → TG'$ by $d/dt (φ(g(t)) = φ_*(g(t))[g'(t)]$, which acts linearly on tangent vectors $g'(t)$ at $g(t)$ and the operation has the signature
$$φ_*(g(t)): T_{g(t)}G → T_{φ(g(t))}G'.$$
I prejudiced the issue by using $G$ to denote a generic differential manifold (one normally uses the letter $M$ to denote manifolds). All of the above applies generically to all manifolds, not just group manifolds. But, if $G$ is also a Lie group, then by definition that means it is a differential manifold which also happens to have group operations on it that are, themselves, differentiable.
My favorite definition - and one of the simplest - is that a Lie group $G$ is a differential manifold containing a distinguished element $e ∈ G$ and a differentiable operation $g,h ∈ G ↦ g/h ∈ G$ such that $g/e = g$, $g/g = e$ and $(g/k)/(h/k) = g/h$, for all $g,h,k ∈ G$. The group operations can be defined from this as $g^{-1} ≡ (g/g)/g [= e/g]$ and $gh ≡ g/h^{-1}$, and you can verify that all the usual properties of groups are satisfied and that the group axioms, themselves, are equivalent to the three quotient axioms above under the correspondence $g/h ↔ g h^{-1}$ (and this equation, in turn, is also a consequence of the three quotient axioms since $g h^{-1} = g/(h^{-1})^{-1}$ and $(h^{-1})^{-1} = e/(e/h) = (h/h)/(e/h) = h/e = h$).
One can define the "translation" maps $L(g) h = g h = R(h) g$, for any $g, h ∈ G$. Their derivatives effectively extend the group operation to the tangent bundle $TG$ by way of the following:
$${d \over dt} (g h(t)) = g h'(t) ≡ L(g)_*(h(t))[h'(t)],$$
$${d \over dt} (g(t) h) = g'(t) h ≡ R(h)_*(g(t))[g'(t)].$$
As algebraic operations, they have the following signatures:
$$\dot g ∈ T_gG ↦ \dot g h ∈ T_{gh}G,$$
$$\dot h ∈ T_hG ↦ g \dot h ∈ T_{gh}G.$$
where $g,h ∈ G$. It satisfies all the expected properties; e.g. $e \dot g = \dot g$, $(g \dot h)k = g(\dot h k)$, $(\dot g h)k = \dot g (hk)$, and so on.
Curiously, mathematicians go out of their way to avoid this economy of notation, so that things get needlessly obscured as a result. The only explanation I can find for that is that they're doing so intentionally for job security - the "Egyptian Scribe" Creed: ("use hieroglyphs, instead of the Phoenician writing, to keep everyone else out of our field"): a form of Academese. "What is Academese?" Thought Co. https://www.thoughtco.com/academese-prose-style-1688963. Or else, the only other explanation I can think of is that there is simply a blind spot in the entire community with respect to the creed "keep it simple".
In fact, the same convention can also be applied more broadly. If a group $G$ acts differentiably on a manifold $M$, then that means one has a differentiable operation
$$g ∈ G, m ∈ M ↦ gm ∈ M$$
such that $em = m, g(hm) = (gh)m$, for all $g,h ∈ G$. Then one can equally well extend the group operation to the tangent bundle $TM$ with the signatures
$$\dot m ∈ T_mM ↦ g \dot m ∈ T_{gm} M$$
for $g ∈ G, m ∈ M$ and even define a map from $TG$ to $TM$ by
$$\dot g  ∈ T_gG ↦ \dot g m ∈ T_{gm} M$$
for $g ∈ G, m ∈ M$. A similar observation holds, if $G$ acts on $M$ to the right.
As we've already seen, this applies to the case where the manifold $M$ is the group $G$, itself: $G$ acts on itself on both the left and the right. So, the group operation applies to both the left and the right on the tangent bundle $TG$ ... which also means that $G$ acts to both the left and right on $TG$.
Sometimes, the right action is taken as a quotient $R(h) g = g/h$, so that the order of operations is preserved; i.e. $R(hk)g = g/(hk) = (g/k)/h = R(h)R(k)g$. I've seen both conventions used. That should be kept in mind. Otherwise, if it is defined as a product on the right, then one would instead have $R(hk) = R(k)R(h)$.
All the expected rules apply, provided one stays within the restrictions laid out by the signatures of the operations. Thus, for instance, we have the same rules for products and quotients as you would for matrix algebra:
$${d \over dt} (g(t) h k(t)) = g'(t) h k(t) + g(t) h k'(t) ∈ T_{g(t)hk(t)}G,$$
$${d \over dt} g(t)^{-1} = -g(t)^{-1} g'(t) g(t)^{-1} ∈ T_{g(t)^{-1}}G.$$
Every group acts on itself in a third way, by operating on the left as a product and at the same time on the right as a quotient. That's the internal automorphism representation
$$Int(g) h = g h g^{-1}.$$
You can verify that is satisfies the required properties $Int(e) g = g$, $Int(gh)k = Int(g)Int(h)k$ to qualify as an action of $G$ on itself.
The differential of this $Ad(g) = Int(g)_*$ is given its own name
$$Ad(g) \dot h = g \dot h g^{-1} ∈ T_{ghg^{-1}}G,$$
and is called the adjoint action of $G$ on $TG$. Curiously, mathematicians don't define it this broadly - not on the entire tangent bundle, but only on $T_eG$. But it can be, and we will do so.
If $g$ and $h$ commute (that is, $gh = hg$), then $T_hG$ is closed under $Ad(g)$: $Ad(g): T_hG → T_hG$. In particular, since $e$ commutes with everyone, then $T_e G$ is closed under $Ad(g)$ and it produces a bona fide operation on $T_e G$:
$$Ad(g) \dot e = g \dot e g^{-1} ∈ T_{geg^{-1}}G = T_eG.$$
The Lie algebra $Lie(G)$ of $G$ is one and the same as the tangent space $Lie(G) ≡
T_eG$ at the identity $e$. Because $Lie(G)$ is closed under $Ad(g)$, with $Ad(g)$ being differentiable, then $Ad(g(s)) \dot e$ forms a differentiable path in $T_eG$ (noting that $T_eG$ is, itself a manifold, as are all tangent spaces). So, one can differentiate $Ad(g(s))$, too
$$Ad_*(g(s))[g'(s)] \dot e = {d \over ds} (g(s) \dot e g(s)^{-1}) ∈ T_eG.$$
This operation also applies more broadly, to tangent spaces $T_hG$ ... provided that $h$ commutes with all of the $g(s)$'es:
$$Ad_*(g(s))[g'(s)] \dot h = {d \over ds} (g(s) \dot h g(s)^{-1}) ∈ T_hG,$$
if $g(s) h = h g(s)$ for all $s$.
If this were a matrix algebra, we would be able to write
$$\begin{align}
‟{d \over ds} (g(s) \dot h g(s)^{-1}) & = g'(s) \dot h g(s)^{-1} - g(s) \dot h g(s)^{-1} g'(s) g(s)^{-1} \\
& = (g'(s) g(s)^{-1}) (g(s) \dot h g(s)^{-1}) - (g(s) \dot h g(s)^{-1}) (g'(s) g(s)^{-1}) \\
& = [g'(s) g(s)^{-1}, g(s) \dot h g(s)^{-1}]”; 
\end{align}$$
note my use of Air Quotes. In particular, if $g(0) = e$, then writing $g'(0) = \dot e$, we would have
$$‟\left({d \over ds} (g(s) \dot h g(s)^{-1})\right)_{s=0} = [\dot e, \dot h]”.$$
But since there is no matrix algebra undergirding $T_eG$ (it was defined as the tangent space of a manifold, not as a matrix algebra), then this correspondence can only be taken as a motivation for defining the Lie bracket in the abstract by way of the equation
$${d \over ds} \left(g(s) \dot e g(s)^{-1}\right) = [g'(s) g(s)^{-1}, g(s) \dot e g(s)^{-1}],$$
for all $\dot e ∈ T_eG$ which would actually apply generally, irrespective of whether $s = 0$ or not, and whether $g(s) = e$ or not. Since both $g'(s) g(s)^{-1} ∈ T_eG$ and $g(s) \dot e g(s)^{-1} ∈ T_eG$, then this is actually an operation on $T_eG$. That the operation is actually well-defined is actually already assured, for the most part, by the way tangent spaces were defined in the first place.
The usual definition assumes $s = 0$ and that $g(0) = e$, so that, with these assumptions, we can just write:
$$\left({d \over ds} (g(s) \dot e g(s)^{-1})\right)_{s=0} = [g'(0), \dot e].$$
Additional Remarks Added On Edit:
Some additional notes on the alternate definitions for Lie algebras:
A vector field $X$ is defined on a manifold $M$ as a differentiable map with the signature $m ∈ M ↦ X(m) ∈ T_mM$.
For each Lie group $G$, and each tangent vector $\dot e ∈ T_eG$, both of these are differentiable vector fields:
$${\dot e}_L: g ∈ G ↦ g \dot e ∈ T_gG,$$
$${\dot e}_R: h ∈ G ↦ \dot e h ∈ T_hG,$$
satisfying the properties
$${\dot e}_L(gh) = (gh) \dot e = g (h \dot e) = g {\dot e}_L(h),$$
$${\dot e}_R(gh) = \dot e (gh) = (\dot e g) h = {\dot e}_R(g) h.$$
Any vector field that is covariant on the left $X(gh) = gX(h)$ can be normalized to the form $X(g) = g X(e)$, with the result being that $X = {X(e)}_L$. The same applies in the other direction: if $X$ is covariant on the right $X(gh) = X(g)h$, then it normalizes to the form $X = {X(e)}_R$.
The actual conversion is done by taking the quotient on the left or right: $g\backslash X(g) = X(e)$ if $X$ is covariant on the left, or $X(h)/h = X(e)$, if it is covariant on the right. The operations themselves $g\backslash (⋯)$ and $(⋯)/h$ have names: Cartan-Maurer forms and can be defined much more broadly than mathematicians usually define them, with the signatures:
$$g\backslash (⋯): T_hG → T_{g\backslash h}G,$$
$$(⋯)/h: T_gG → T_{g/h}G.$$
As functions of $g$, when restricted to the signature $T_gG → T_eG$, they map the tangent bundle $TG$, itself, to the Lie algebra $T_eG$.
So, by construction, there is one vector field of each type, for each tangent vector at $T_eG$, so either set of vector fields can be taken as a proxy for $T_eG$, itself. The main significance of this remark is that there is already a way to define Lie brackets for vector fields on manifolds; and the vector fields, with this Lie bracket, can be taken as an alternate way to define the Lie algebra for the Lie group $G$.
The manifold-theoretic Lie bracket arises as a result of the fact that the process of going from curve to tangent vector (by differentiating to get $m'(t) ∈ T_{m(t)}M$ on a differential manifold $M$) can be reversed, with vector fields, by solving the differential equation $d/dt(x(t)) = X(x(t))$. This produces a family of curves, called the flow of $X$. Once you convert vector fields to flows, then you have curves.
A vector field can effectively be defined by the way it acts on real-valued functions over the manifold in such a way that if $f: M → ℝ$, then $d/dt (f(x(t))) = x'(t) [f] = X(x(t)) [f]$, where $x(t)$ is the flow of $X$.
In general, for any differentiable path $x(t)$ on a differential manifold $M$, one has the operation defined by $x'(t)[f] = d/dt (f(x(t)))$. That extra detail is actually how tangent vectors on a differential manifold are actually defined and constructed in the first place.
The Lie bracket $[X,Y]$ of two vector fields $X$ and $Y$ is defined by $[X,Y][f] = X[Y[f]] - Y[X[f]]$, which serves to define $[X,Y]$ as a vector field.
The Lie algebraic definition of the Lie bracket matches the manifold-theoretic definition of the Lie bracket for the corresponding vector field: $[{\dot e}_0, {\dot e}_1]_L = [{\dot e}_{0L}, {\dot e}_{1L}]$. The sign is reversed for the other direction, $[{\dot e}_0, {\dot e}_1]_R = -[{\dot e}_{0R}, {\dot e}_{1R}]$, so there is a preference for vector fields covariant on the left over fields covariant on the right, when defining Lie algebras in terms of such vector fields.
