Lie algebra Representation of the group $(\mathbb{R},+)$

Question

I am reading Peter Woit's book Quantum Theory, Groups and Representations, section 10.1 (a similar version can be found here). I am reading the part where he introduces the representation of $(\mathbb{R},+)$ on $L^2(\mathbb{R}).$ He said the following:

The simplest case is the representation induced on functions on $\mathbb{R}$ by the action of $\mathbb{R}$ on itself by translation. Here, $a\in \mathbb{R}$ acts on $q\in \mathbb{R}$ (where $q$ is a coordinate on $\mathbb{R}$) by $$q\to a\cdot q=q+a,$$ which induces a representation $\pi:\mathbb{R}\to GL(L^2(\mathbb{R}))$ given by $$\pi(g)f(q)=f(g^{-1}\cdot q)$$ for which this case will be $$\pi(a)f(q)=f(q-a).$$ To get the Lie algebra version of this representation, the above can be differentiated, finding $$\pi'(a)=-a\frac{d}{dq}.$$

My Question: I don't see how to go from $\pi(a)f(q)=f(q-a)$ to $\pi'(a)=-a\frac{d}{dq}.$ I would appreciate if someone can explain it to me. Here is what I have done so far:

By definition, the derivative $\pi'$ of the induced Lie algebra representation is related to the Lie group representation via $$\pi'(X)=\frac{d}{dt}\pi(e^{tX})|_{t=0}.$$

Therefore, we have that

$$ \pi'(a)f(q)=\frac{d}{dt}\pi(e^{ta})f(q)|_{t=0}=\frac{d}{dt}f((-e^{ta})\cdot q)|_{t=0}=\frac{d}{dt}f(q-e^{ta})|_{t=0}=[f'(q-e^{ta})(-ae^{ta})]_{t=0}=-af'(q-1). $$

So I didn't get the expected answer. Instead, I get a shift of the derivative $f'(q-1)$, but it should have been $f'(q)$.

Another method I try to use is by defining $$\pi'(a)=\lim_{h\to 0}\frac{\pi(a+h)-\pi(a)}{h}.$$ But in the end I get $\pi'(a)f(q)=-f'(q-a).$ Another different answer, which is very weird. So I must have made some silly mistakes in my work.

Answer 1

You must be careful - remember that the Lie algebra is given by the tangent space to the Lie group at the identity element. The identity element of $G=(\mathbb R,+)$ is not $1\in \mathbb R$ but rather $0\in \mathbb R$. There are two ways to study this somewhat odd case clearly. We may either implement $G$ as a matrix group, or we may treat it as a smooth manifold.

The former is simpler. Define $$G := \left\{\pmatrix{1&a\\0&1}\in \mathbb R_{2\times 2}\ \bigg | \ a\in \mathbb R\right\}\simeq (\mathbb R,+)$$ In this form, the group operation is given by matrix multiplication, since $$ \pmatrix{1&a\\0&1} \pmatrix{1&b\\0&1} = \pmatrix{1&a+b\\0&1}$$ Letting $a$ be infinitesimal, we find that the Lie algebra takes the form $$\mathfrak g := \left\{\pmatrix{0&a\\0&0}\ \bigg| \ a\in \mathbb R\right\} \simeq (\mathbb R,+)$$ The exponential map is then straightforwardly given by $$\exp\left(\pmatrix{0&a\\0&0}\right)= \pmatrix{1&a\\0&1}$$ Observe that both $G$ and $\mathfrak g$ are isomorphic to $(\mathbb R,+)$ - the former as a group, and the latter as a vector space - and that $G\ni \pmatrix{1&a\\0&1}$ and $\mathfrak g\ni \pmatrix{0&a\\0&0}$ may be identified with $a\in \mathbb R$. This suggests - somewhat counterintuitively - that we might write $\exp(a) = a$, where the exponential map is (clearly) not the familiar exponential from elementary analysis.

Answer 2

$\pi(a) f(q)=f(q- a) \approx f(q) - f'(q) a$

$\pi(a) = e^{-a \frac{\partial}{\partial q}} \approx 1- a\frac{\partial}{\partial q}$

$\pi'(a) |_{a=0} = - \frac{\partial}{\partial q} e^{-a \frac{\partial}{\partial q}}|_{a=0}\approx -\frac{\partial}{\partial q}$

Physics From Symmetry. Jakob Schwictenberg. Page 40.

Physics From Finance. Jakob Schwictenberg. Page 160.