*at*the identity, we could smear it across the group to form a

**vector field**, resolving questions of whether our tangent space "really needs to be" at the identity (the exponential map in matrix representation only exists in the traditional form if we're talking about tangent vectors at the identity, but we're free to write down the Lie algebra in this way).

But not every vector field is a valid element of the Lie algebra. We need the vector field to be "

**constant**" across the manifold in some sense so that that constant vector it equals is the tangent-space-at-the-identity element it corresponds to. But what exactly do we mean by "constant" on a Lie Group?

In the case of the unit circle in the complex plane, we have an idea of what we want -- the vector field $T(M)$ is constant over the group if it is determined by the value at the identity as $T(M)=MT(0)$.

Is this preserved in the matrix representation of the group? Well, yes, because the correspondence between complex numbers and spiral matrices is a homomorphism. We can use this as a motivation to define the condition for a vector field to be a Lie algebra on a matrix Lie group -- it needs to be a

**left-invariant vector field**, i.e. we need that the value of the vector field determined as $T(M)=MT(0)$.

Why left-invariant? Why not right-invariant? Why matrix multiplication at all? The choices made here are certainly arbitrary to some extent. When we study

**abstract lie algebra**, we'll just have "left-multiplication by $M$" being replaced by a group action and the usage of matrix multiplication is a**choice of representation**. In the context of abstract Lie algebra, the "left-multiplication by $M$ we're interested in is really the*derivative*of the group homomorphism $M:G\to G$, which is a linear map between the tangent spaces at $I$ and $M$. You can show that this map is represented by matrix left-multiplication given a matrix representation (i.e. letting the group be $GL(n,\mathbb{C})$).Ok, why did we just do that? Why did we upgrade our tangent vectors to vector fields? If it wasn't obvious already, the

**noncommutativity**of a Lie group is "the" feature of importance in a Lie group, at least in some neighbourhood of the identity (we will later find out exactly the kind of features that aren't determined by just the Lie bracket -- the important keywords here are

**connected**and

**compact**) -- if the Lie group is commutative, then the Lie algebra is just a vector space with no additional structure, and the Lie group is a "basically unique" choice.

In our discussions of noncommutativity in the last article, we repeatedly referred to

*flowing along a vector*-- the nature of noncommutativity is inherently "dynamical" in this sense. So we need to talk about

*differentiating along the corresponding vector field to a tangent vector*.

So let's upgrade our vector fields to derivative operators, or

**derivations**$D$. These are operators on functions $f:G\to \mathbb{R}$ that tell you the derivative of $f$ in the direction of the vector field -- the left-invariant ones are a certain generalisation of the directional derivative operators.

Well, what exactly is a derivation? On Euclidean space, directional derivatives can be imagined as stuff of the form $f\mapsto\vec{v}\cdot\nabla f$ -- but this requires the concept of a dot product which is quite weird within the context of matrix groups. But if you try to work this out on the unit circle (do it!), you might get an idea: we can define a

**curve**$\gamma:\mathbb{R}\to G$ passing through a point and consider:

$$f\mapsto(f\circ \gamma)'(t)$$

At the point and you get precisely the directional derivative in the direction $\gamma'(t)$ (show that this is right in Euclidean space, and make sure you understand why it is right/makes sense -- it's the chain rule, and a certain analogy exists to projecting matrices onto subspaces in linear algebra). And if we just want tangent vectors at the identity, we can just consider the operation $f\mapsto(f\circ \gamma)'(0)$.

OK. Let's try to "

**abstract out**" the properties of a derivation $D$, i.e. something that just allows us to define what a derivation is, abstractly, that is equivalent to being an operator of the above form.

What makes an operator a directional derivative? Certainly it must be a linear operator -- but not every linear operator is a directional derivative. The key idea behind a directional derivative is that $D(f(x))$ is

**determined in a specific way by $D(x)$, the rate at which $x$ changes**in the specified direction.

How do we use this? Well, if you think about it a little bit, we can restrict $f$ to be analytic -- so we need:

- $D(x)$ predicts $D(x^n)$ in the right way -- this is ensured by the
**product rule**-- $D(fg)=f\ Dg + g\ Df$. - $D(x^n)$ for all $n$ predicts $D(a_0+a_1x+a_2x^2+\ldots)$ in the right way -- this is ensured by
**linearity**.

If anyone can motivate the definition of a derivation without restricting to analytic functions, tell me.

An operator that satisfies these two properties is called a

**derivation**-- one can prove additional properties from these axioms fairly easily, e.g. $D(c)=0$ for constant $c$, etc.

Let's think about why this whole construction above makes sense.

Let $G$ be the group of translations of $\mathbb{R}$ -- one can parameterise them by the translated distance as $\Delta(p)$ with composition given by $\Delta(p)\Delta(q)=\Delta(p+q)$. Well, this is isomorphic to the additive group on the reals, and in turn to the multiplicative positive real numbers. We can consider the group to be acting on real analytic functions by translations of the domain: $\Delta_pf(x):=f(x+p)$ The Lie algebra is just spanned by the derivative of $\Delta(p)$ at the identity, that is:

$$\Delta '(0) = \lim\limits_{h \to 0} \frac{{\Delta (h) - 1}}{h} = \frac{d}{{dx}}$$

And our Lie algebra members are all real multiples of $d/dx$ -- these are precisely the directional derivatives on $\mathbb{R}$. Similar constructions can be made on $\mathbb{R}^n$, or a general automorphism group.

So we see that the "derivations" construction of the Lie algebra actually are

**the tangent vectors on the Lie group identified as the automorphism group of some object**. If you've ever done some differential geometry, this gives you the motivation for treating partial derivatives as basis vectors.

Our discussion of derivations so far works both for derivations (general vector fields on the manifold) and

$$p\frac{d(f\cdot g)}{dx}=g\cdot p\frac{df}{dx}+f\cdot p\frac{dg}{dx}$$

And why shouldn't it? It corresponds to a vector field all right -- $xe_x$. But this is not a

**point-derivations**(basically tangent vectors at a specific point). Under the first interpretations, though, we're**not actually interested in all derivations**, only the left-invariant ones. For example, in the example above, an operation of the form of $p(x)\frac{d}{dx}$ is linear and satisfies the product rule:$$p\frac{d(f\cdot g)}{dx}=g\cdot p\frac{df}{dx}+f\cdot p\frac{dg}{dx}$$

And why shouldn't it? It corresponds to a vector field all right -- $xe_x$. But this is not a

*left-invariant vector field*.
Interpret the

**Taylor series as the exponential map**from the Lie algebra to the Lie group! Make the "similar construction" in the multivariate case ($\mathbb{R}^n$) and interpret the**multivariate taylor series**as an exponential map -- i.e. that $\Delta=\exp\nabla$
$$[D_1,D_2](fg)=f[D_1,D_2]g+g[D_2,D_1]f$$

I.e. that the Lie Bracket of two derivations is also a derivation. Check that the above is correct by expanding stuff out and using the product rule for $D_1$ and $D_2$.

There's another way that derivations can be used to show closure under the Lie Bracket, which shows more closely the connection to the product rule for the second derivative discussed in the previous article.

One might wonder if, like the directional derivative at the identity in the $c'(0)$ direction is given by $(f\circ c)'(0)$, the directional derivative at the identity in the $c''(0)$ direction may be given as $(f\circ c)''(0)$. Well, in general:

$$(f\circ c)''(t)=c''(t)\cdot\nabla f(t)+c'(t)\frac{d}{dt}\nabla f(t)$$

Which since $c'(0)=0$, at $t=0$ is simply equal to the first term, the directional derivative in the $c''(0)$ direction. So we just need to show that $f\mapsto (f\circ c)''(0)$ is a derivation. This follows from the

One might wonder if, like the directional derivative at the identity in the $c'(0)$ direction is given by $(f\circ c)'(0)$, the directional derivative at the identity in the $c''(0)$ direction may be given as $(f\circ c)''(0)$. Well, in general:

$$(f\circ c)''(t)=c''(t)\cdot\nabla f(t)+c'(t)\frac{d}{dt}\nabla f(t)$$

Which since $c'(0)=0$, at $t=0$ is simply equal to the first term, the directional derivative in the $c''(0)$ direction. So we just need to show that $f\mapsto (f\circ c)''(0)$ is a derivation. This follows from the

**Leibniz rule for the second derivative**, and the fact that the first derivative of $c$ is zero.
OK, one more thing before we actually do something useful -- something we haven't done before in other ways.

This is an

$$D=\phi^{h}D\phi^{-h}$$

This is an

**extended pitfall prevention**, because I fell into this pit myself. When thinking about left-invariance of a vector field $D$, I formulated the idea in my head this way: the idea is that under $D$, we should get the same result if we differentiate (derivate?) $f$ at 0 or if we translate it forward by $x$ and derivate it at $x$. i.e. where $\phi^h$ represents the translation $f(x)\mapsto f(x-h)$, we want:$$D=\phi^{h}D\phi^{-h}$$

(

But it's wrong. How do we know that? Well, let's consider the group action $\phi^{-h}D\phi^h$ -- certainly at $h=0$, it's the identity, so let's differentiate it (against $h$) at 0. We get, where $d\phi_0$ is the derivative of $\phi$ at 0:

$$[d\phi_0, D]$$

Which isn't zero. So my argument must be wrong -- I must have assumed abelian-ness somehow.

Here's the problem: the final left-multiplication by $\phi^h$ is fine -- it just brings the derived function back to the origin, but "translating the function forward and then differentiating it" messes things up when the direction you're differentiating in doesn't commute with the direction of translation. Draw some pictures of curved surfaces to convince yourselves of this.

So left-multiplication determines a sort of "parallel transport" on the Lie Group, while right-multiplication is an "alternative" way to compare vectors in different tangent spaces, and its disagreement with left-multiplication determines the non-commutativity of the group. Well, this choice of left-multiplication vs right-multiplication is really a convention, arising from the choice of representation.

*THIS IS WRONG!*This is a pitfall prevention, not an actual result!) And I looked at some simple Abelian cases, like the additive real group and the circle group and thought this was clearly true.But it's wrong. How do we know that? Well, let's consider the group action $\phi^{-h}D\phi^h$ -- certainly at $h=0$, it's the identity, so let's differentiate it (against $h$) at 0. We get, where $d\phi_0$ is the derivative of $\phi$ at 0:

$$[d\phi_0, D]$$

Which isn't zero. So my argument must be wrong -- I must have assumed abelian-ness somehow.

Here's the problem: the final left-multiplication by $\phi^h$ is fine -- it just brings the derived function back to the origin, but "translating the function forward and then differentiating it" messes things up when the direction you're differentiating in doesn't commute with the direction of translation. Draw some pictures of curved surfaces to convince yourselves of this.

So left-multiplication determines a sort of "parallel transport" on the Lie Group, while right-multiplication is an "alternative" way to compare vectors in different tangent spaces, and its disagreement with left-multiplication determines the non-commutativity of the group. Well, this choice of left-multiplication vs right-multiplication is really a convention, arising from the choice of representation.

OK, the useful thing: Suppose we're interested in "nested Lie brackets" $[X,[Y,Z]]$. We're talking about conjugating $[Y,Z]$ as $\phi^p[Y,Z]\phi^{-p}$ where $d\phi_0=X$ so that to first-order in $p$:

$$\phi^p[Y,Z]\phi^{-p}=[Y,Z]+p[X,[Y,Z]]$$

Since conjugation is a homomorphism, we can also write:

$$\begin{align}

\phi^p[Y,Z]\phi^{-p} &= [\phi^pY\phi^{-p},\phi^pZ\phi^{-p}] \\

&= [Y+p[X,Y],Z+p[X,Z]] \\

&= [Y,Z] + p([Y,[X,Z]]+[[X,Y],Z])\\

\Rightarrow [X,[Y,Z]]&=[Y,[X,Z]]+[[X,Y],Z]

\end{align}$$

Now, couldn't we have just have proven this by expanding everything out as commutators? Sure, but this provides more insight as to what's going on -- you might notice the resemblance to the product rule. Indeed, this identity --

**the Jacobi identity**-- is perhaps best stated as:

"

**A derivation $X$ acts through the Lie Bracket as a derivation on the space of derivations where "multiplication" is given by the Lie Bracket.**"

In this sense, it's actually quite expected -- it results from the fact that the Lie Bracket is a bilinear operator obtained from

**differentiating a group symmetry, conjugation**-- this mandates that it is a derivation.

As it turns out, the Jacobi identity, along with the antisymmetry and the bilinearity, determines the Lie Algebra -- it is enough to "abstract out" the properties of a Lie Algebra. Why? This is something we will see over several articles, which will then allow us to motivate abstract Lie algebra.

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