How do I know that a group generator really is from that group? I've asked the question here but I'm actually not that satisfied with an answer.
I have 3 generators that are (at least from what I read in every article I find) generators of $SL(2,\mathbb{R})$ group. But how do I know that they (the $\tilde{J}_i$ ones) are indeed generators of $SL(2,\mathbb{R})$?
Sure one can regard it as a change of basis (although I couldn't check the $M T_i M^{-1}=\tilde{T}_i$ rule with a given matrices, not even sure that's possible, since on one side I have $2\times 2$ matrix, and on the other a $4\times 1$ matrix (vector)).
But in general. How do I know that certain generator is indeed a generator of some specific group. Because, if it satisfies the rules of Lie algebra, that just means it forms a Lie algebra, not what that algebra is.
 A: Okay. It took me a while to reorder all my terms to make it the more obvious but here I show that the two algebras have the same Lie-structure
Your Lie algebras is at heart a Vector space and  the $X,Y,Z$ matrices for a basis for the vector space. We know that we can just go ahead and form a new basis for the vector space by means of linear combinations of this basis and the one I am suggesting we transfer to is
$$J_0 = Y - Z, \qquad J_1 = X, \qquad J_2 = Y + Z$$
It's still the same Lie algebra since we let the Lie-bracket be as before and we find the brackets of these basis matrices to be
$$[J_0, J_1] = [Y - Z, X] = -2Z - 2Y = -2J_2$$
$$[J_0, J_2] = [Y - Z, Y + Z] = 2X = 2J_1$$
$$[J_1, J_2] = [X, Y  + Z] = 2Y - 2Z = 2J_0$$
From which we easily compare with the $\tilde{J}_i$-brackets and find that the structure is precisely the same. The linear map $f$ which maps matrices to differential operators according to
$$f : J_i \to \tilde{J_i}$$
could be used to form a Lie-algebra-homomorphism between these forms of $\mathfrak{sl}(2)$ so we find that they are indeed the same algebra. 
(This is the first proof I have performed of this type and I am still quite new to this field so please check it for errors)
