# (Physics version of) Taylor expansion. In the the context of deriving a Lie groups generators (a Lie algebra from a Lie group)

Statement which I'm confused about:

"Consider some n-dimensional Lie group whose elements depend on a set of parameters $$\alpha = (\alpha_1 ... \alpha_n)$$, such that $$g(0) = e$$ with e as the identity, and that had a d-dimensional representation $$D(\alpha)=D(g( \alpha),$$ such that $$D(0)= \mathbb{1}_{d \times d}$$. Then in some small neighborhood of $$\mathbb{1}$$, we can expand $$D(\alpha)$$ as, $$D(d\alpha) = \mathbb{1} + i d \alpha_i X^i,$$ where $$X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}$$."

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Using this terminalology on the Lie case:

$$\begin{eqnarray} D(0+d\alpha) &=& D(0) + \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) (-i) \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) X^i d\alpha \end{eqnarray}$$

is this correct? Also, why is the "taking the derivative at $$\alpha=0$$ important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?

• Please do not post images of texts you want to quote, but type it out instead so it can be indexed by search engines. For formulae, use MathJax instead. – ACuriousMind Oct 13 '19 at 9:33

1. For matrix groups - i.e. those $$G$$ embeddable in $$n$$-by-$$n$$-matrices $$\mathbb{R}^{n\times n}$$ - this is really just Taylor expansion of a function $$D: \mathbb{R}^n \to \mathbb{R}^{n\times n}$$. There's nothing special about it and no trick to it (except the factors of $$\mathrm{i}$$ to make the $$X$$ be Hermitian).
That we're expanding at $$0$$ - corresponding to $$D(0)$$ being the identity - is simply because for an arbitrary Lie group we don't know what $$D(\alpha)$$ for $$\alpha\neq 0$$ is beyond "some element". We could expand around that, but it wouldn't be very useful.
2. The $$\delta \alpha$$ in your quote is supposed to be $$\delta \alpha = \frac{\alpha}{N}$$. $$\mathrm{e}^{x} = \lim_{n\to\infty} (1 + \frac{x}{n})^n$$ should be a more familiar limit to you.
• Can you expand on why $\delta \alpha = \frac{\alpha}{N}$? We had that $\delta \alpha$ is an infinitesimal quantity, so it seems off-putting to write it as $\frac{\alpha}{N}$. When N is not approaching infinity, $\delta \alpha$ is not infinitesimal. Am I missing something? – Lopey Tall Oct 13 '19 at 13:23
• @LopeyTall In that part of the text (everything after "we can translate through $\delta a$"), $\delta \alpha$ is indeed not an infinitesimal. It's crappy notation on the authors' part to reuse $\delta \alpha$ there. – ACuriousMind Oct 13 '19 at 13:34
• Checking my prof's Lie groups lecture notes, he indeed says "where $d \alpha$ denotes an infinitesimal $\alpha$." This is before he goes on to make the substitution you mention ($d \alpha=\frac{a}{N})$ to make the limit more evident to the reader. Can you explain why you think the $\delta \alpha$ is not infinitesimal? – Lopey Tall Oct 13 '19 at 17:46