Date: [[2025-11-17]] I just saw this video and got my mind blown: https://youtu.be/xdwuyXrObSA?si=SpbnSHMchUSpc6Z2. Some thoughts I have that build on it (no need to watch the video): First: - You know [[Newton's Binomial Theorem]], right? e.g. $(x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3$ - Well consider this: $\frac{d^3}{dx^3} f(x) g(x) = f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + g'''(x)f(x)$ - Surprising! There exist combinatorial explanations for this, but is there something deeper here? Perhaps -- perhaps a derivative operator D can be affixed onto an algebra to make a [[Lie Algebra]]? - Consider the primitive of [[Lie Algebra]], the [[Commutator]]: $[x, y] = xy - yx$ - With $D$ as an element of the [[Lie Algebra]], for some function $f$ you could write: $[D, f] = Df - fD$ - That may seem meaningless for now. But let's pretend that $[D, f]$ is a notational standin for the derivative of $f$, and see what happens - Let's write the product rule: $[D, fg] = Dfg - fgD = Dfg - (fDg - fDg) - fgD = [D, f]g + f[D, g]$ - We can then try a second derivative: $[D, [D, fg]] = [D, [D, f]g + f[D, g]] = [D, [D, f]] + 2[D, f][D, g] + [D, [D, g]]$ - In general, treating $D$ and the [[Commutator]] in this way lets you derive pretty much all the algebraic properties of the derivative > [!NOTE] WIP > The above is about as far as I have gotten with this. I want to somehow complete the following incomplete chain of logic, in order to fully figure this out. The end goal is presumably formal justification for all sorts of operations related to manipulating $\frac{d}{dx}$, but there are still a lot of gaps in my understanding # WIP In this framework, both $D$ and $f$ are simply elements of a [[Lie Algebra]], and nothing we have defined thus far makes $D$ fundamentally different from $f$ - So we could equally look at $[f, D]$, which is equivalent to $-[D, f]$ - We have been treating $f$ abstractly as a function, that could potentially take an input like $f(x)$. What then is $D(x)$? - Let's try to plug in a polynomial $f(x) = x^2 + 2x + 1$. We know then that in principle $[D, f] = 2x + 2$ in some fashion: $f(x) = x^2 + 2x + 1 \implies 2x + 2 = Df - fD$ - Similarly, it should be the case that $[D, [D, f]] = 2$, so we obtain $[D, [D, f]] = [D, (Df - fD)] = DDf - 2DfD + fDD = 2$ We know the other big part of [[Lie Algebra]] is the [[Exponential Map]]. What then does $e^D$ mean? - Recall the [[BCH Formula]]: to solve for $z$ in $e^x e^y = e^z$ in a noncommutative [[Lie Algebra]], you write: $z = x + y + \frac{1}{2}[x, y] + \frac{1}{12}([x, [x, y]] + [y, [y, x]]) + \cdots$ - Let's consider then the first order of $? = \log(e^D e^f)$: $? = D + f + \frac{1}{2}[D, f]$