Let $G$ be an algebraic group (although the question is the dot point at the bottom, and I suppose you can consider Lie groups, where $K[G]$ in the Lie group setting are just functions $f:G\to K$ on the Lie group?)
I am trying to understand left invariant derivations of functions.
So let $\delta:K[G]\to K[G]$ be a derivation, and let $\lambda_x$ be a left translation of functions, meaning that:$$\lambda_x f(y)=f(x^{-1}y), \quad(x,y\in G, f\in K[G]).$$
Left invariant derivation, means that $\lambda_x \delta = \delta \lambda_x$, for each $x\in G$.
I think they mean composition, so that we just ask: $$\begin{matrix}K[G]&\stackrel{\lambda_x}{\to}&K[G]\\\delta\downarrow&&\downarrow \delta\\K[G]&\stackrel{\lambda_x}{\to}&K[G]\end{matrix}$$
Then I want to think of an example, so let $B\subset \text{SL}_2$ be a Borel subgroup, where we write $K[X_1,X_1^{-1},X_2]$ for the associated algebra. Let $\delta = \frac{\partial}{\partial X_1}$.
Then I want to see how this would fail to commute (if it does), so I look at $f= X_1^2+X_1X_2$ and then $\delta f = 2X_1^2 + X_2$ and $\lambda_x\delta(f)$ just multiplies any element of $G$ by $x^{-1}$ first, before evaluating $\delta(f)$. This seems fine. (Going down the left and bottom of the diagram)
But going the other way, how do I see $\delta(\lambda_x f)$? Say I let $$x=\begin{bmatrix}a&b\\0&a^{-1}\end{bmatrix}, y= \begin{bmatrix}t&z\\0&t^{-1}\end{bmatrix}\implies \lambda_x f(y)=X_1^2+X_1X_2(\begin{bmatrix}a^{-1}t&a^{-1}z-bt^{-1}\\0&at^{-1}\end{bmatrix})=a^{-2}t^2+a^{-2}tz-a^{-1}b$$ then how would I have thought of $\delta(\lambda_x f)(y)$? I get that $\delta\lambda_x f: G\to K$, and so I cannot apply $\delta$ to $\lambda_x f(y)$, (i.e. when already evaluated) but I would suspect this gives me insight?
- How do I see $\delta(\lambda_x f)$?
(Sorry about the diagram, I couldn't get Tikzcd to work on here for some reason - maybe there is a trick on this website?)
You've calculated $\lambda_x f(y) = a^{-2} t^2 + a^{-2} t z - a^{-1} b$, and since $y = \begin{pmatrix} t & z \\ 0 & t^{-1} \end{pmatrix}$, you can directly read off what $\lambda_x f$ is by just replacing the $t$ and $z$ variables with the coordinate functions: $$\lambda_x f = a^{-2} X_1^2 + a^{-2} X_1 X_2 - a^{-1}b$$ Now it should be easy to apply the derivation and check that $\delta$ is indeed not a left-invariant derivation.
In general, since $\lambda_x: K[G] \to K[G]$ is a $K$-algebra homomorphism, you only need to check what it does on generators, which you can do using the same method as above, and get $$ \begin{aligned} \lambda_x X_1 &= a^{-1} X_1 \\ \lambda_x X_2 &= a^{-1} X_2 - b X_1^{-1} \\ \lambda_x X_1^{-1} &= a X_1^{-1} \end{aligned} $$
Although I think that in general this is a cumbersome way of actually discovering the Lie algebra of $G$, and it's better to look at the tangent space at the identity, and use this to figure out what the bracket should be.