A Unitary representations of Lie group from Lie algebra

[leo.]

TL;DR

In Quantum Mechanics a symmetry group acts upon the Hilbert space of a quantum system by unitary representations. Usually, though, one studies the representations of the Lie algebra to later get the unitary representations of the group out of the algebra. I want to understand the general procedure of getting the unitary representation of a Lie group out of the hermitian representation of its Lie algebra.

In Quantum Mechanics, by Wigner's theorem, a symmetry can be represented either by a unitary linear or antiunitary antilinear operator on the Hilbert space of states $\cal H$. If $G$ is then a Lie group of symmetries, for each $T\in G$ we have some $U(T)$ acting on the Hilbert space and the association $T\mapsto U(T)$ is, in fact, a projective representation of the group $G$. I am aware that such an issue can be dealt with by passing to the universal cover of $G$, so for the moment let us focus on the case on which $U(T)$ is a de facto representation, obeying $$U(T)U(T′)=U(TT′).\tag{1}$$ Now take $\{\theta^a\}$ coordinates on a neighborhood of the identity $e\in G$ with $\theta^a(e)=0$. Let further $T(\theta)$ be the group element with coordinates $\theta^a$. Now expanding $U(T(\theta))$ to first order in $\theta$ we have $$U(T(\theta))=1+it_a\theta^a + O(\theta^2)\tag{2},$$
where we have defined these $t_a$ operators. One can show that the unitary condition on $U(T(\theta))$ corresponds to the $t_a$ being Hermitian operators. It so happens that the $t_a$ obey the commutation relations of the Lie algebra $\mathfrak{g}\equiv T_e G$ of $G$ defined as the tangent space at the identity. In that sense, if we define $$\pi (X) = X^a t_a,\quad \forall X = X^a\frac{\partial}{\partial \theta^a}\bigg|_e \in \mathfrak{g},\tag{3}$$ then $\pi$ is a Hermitian representation of the Lie algebra, derived from the unitary representation of the Lie group $U$.

In that sense if we know the group unitary representation $U$ we find the Lie algebra hermitian representation $\pi$ essentially by expanding at first order around the identity of the group. I'm interested in the reverse of this: we know the Lie algebra hermitian representation $\pi$ and we wish to find $U$ from which it arises.

In the book "The Quantum Theory of Fields, Vol. 1", Chapter 2, Appendix B, Weinberg tries to reconstruct the $U(T(\theta))$ out of the $t_a$ by solving a differential equation which determines the $U$ along a path $\Theta^a(s)$ starting at the origin and discussing the dependence on the choice of path. The differential equation is Eq. (2.B.2) in his book: $$\dfrac{d}{ds}U_\theta(s)=it_aU_\theta(s) h^a_{\phantom{a}b}(\Theta_\theta(s))\dfrac{d\Theta^b_\theta(s)}{ds},\tag{4}$$ where $U_\theta(s)$ is just $U(T(\Theta(s)))$ and $h^a_{\phantom{a}b}$ is defined to be
$$[h^{-1}]^a_{\phantom{a}b}(\theta)=\left[\dfrac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0}.\tag{5}$$
I don't understand, though, why solving this differential equation is the right way to get the $U(T(\theta))$ out of the $t_a$.

So in summary: given a unitary representation of a Lie group we get a hermitian representation of the Lie algebra. If we know, instead, the representation of the Lie algebra, how do we find out the Lie group representation? In other words, how to get $U(T(\theta))$ out of the $t_a$? Why Weinberg's differential equation (2.B.2) is what answers this question?

 

[fresh_42]

I don't understand enough from the language physicists use, but the mathematical answer to

I want to understand the general procedure of getting the unitary representation of a Lie group out of the hermitian representation of its Lie algebra.

is, that the following diagram commutes:
\begin{equation*}
\begin{aligned}
G &\stackrel{\operatorname{\varphi}}{\longrightarrow} GL(V) \\
\exp \uparrow & \quad \quad \uparrow \exp \\
\mathfrak{g} &\stackrel{\operatorname{D\varphi}}{\longrightarrow} \mathfrak{gl}(V)
\end{aligned}
\end{equation*}
So all representations $D\varphi$ on the Lie algebra level can be lifted to the Lie group level via the exponential function. It is the same idea which is used to solve differential equations. The most common example is $\varphi = \operatorname{Ad}$ with $D\varphi = \operatorname{\mathfrak{ad}}$.

 

[leo.]

This seems to be what Robert Ticciati does on the book "QFT for Mathematicians". If I understand, given the Lie algebra representation $D : \mathfrak{g}\to \mathfrak{gl}({\cal H})$ he tentatively defines $U:G\to GL({\cal H})$ to be $$U(\exp tX) = \exp tD(X),\tag{1}$$ on which the $\exp$ on the LHS is the Lie group exponential and on the RHS is the hilbert space exponential.

In the case that $G$ is connected we know that every element of $G$ is a product of exponentials, so provided (1) defines a single-valued function, it determines $U$ completely. I believe that is what you mean, right?

Still, I would like to understand the Physicist approach, which seems to be solving Eq. (4) in the OP, which is the same as Eq. (2.B.2) in Weinberg's "The Quantum Theory of Fields, Vol. 1". I've been thinking for a few days and yet couldn't understand why this is the right approach.

I think the main difference with the approach using the exponential is that in the approach via the exponential map, we are considering a specific path on $G$, namely $\exp tX$ for some $X\in \mathfrak{g}$, while in Weinberg's approach one is considering a general path $\gamma : [0,1]\to G$ with $\gamma(0)=e$, whose chart representative is $\Theta^a(s) = \theta^a\circ \gamma(s)$.

 

[fresh_42]

This was of course an answer which was destillated to the crucial point. Varadarajan spends an entire chapter with more than 100 pages on the subject. Special calculations work with flows and their integral curves. The analogon with solving differential equations isn't by chance. We have an equation $y'=f(x)$ and investigate $y=e^{g(x)}$ where the initial conditions at the end determine the specific path our solution has on the given vector field. The diagram shows the underlying principle, specific flows are related to specific solutions. The point of evaluation is the neutral group element, which translates to the identity matrix in the representation and the zero matrix on the tangent space. I like to stress the evaluation point ($t=0$) because it is essential for calculations. Your equation (4) looks like a flow on a vector field.

I would compute a specific example. $SU(2)$ with the Pauli matrices is small enough to be done manually.
You can find the corresponding theorem about its (finite dimensional) Lie algebra representations $- \;\mathfrak{sl}(2)\;-$ here (7.1):
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
The general situation is described in:
https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/#toggle-id-1
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/
but admittedly from the mathematical point of view.

 

[leo.]

There is an important point about Eq. (4) in the OP that I forgot to mention when starting the thread, which is the definition of the $f(\bar{\theta},\theta)$ map appearing in Eq. (5) when defining the matrix $h^a_{\phantom{a}b}(\theta)$. It is the coordinate representation of the group multiplication. In other words, letting $T(\theta)$ be the group element with coordinates $\theta$, we have $$T(\bar{\theta})T(\theta)=T(f(\bar{\theta},\theta)).$$ In that case $[h^{-1}]^a_{\phantom{a}b}(\theta)$ are the components of the differential at the identity of the right translation by $T(\theta)$: $$[h^{-1}]^a_{\phantom{a}b}(\theta)=\dfrac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}\bigg|_{\bar{\theta}=0}.$$

Your equation (4) looks like a flow on a vector field.

I have tried to pursue this line of thought. For that I've tried to write the differential equation for $\gamma(s) = \exp(sX)$ in this notation to try to understand better what is going on. This should be the integral curve of the right-invariant vector field whose value at the identity is $X$. This means the problem to be studied is $$\begin{cases}\gamma'(s)&=&[R_{\gamma(s)}]_{\ast e}X,\\ \gamma(0)&=&e.\end{cases}$$ Now, in Weinberg's coordinates right translation by $T(\theta)$ takes $\bar{\theta}\mapsto f(\bar{\theta},\theta)$ where the second entry is fixed. This implies that the matrix of the pushforward in the differential equation is just Weinberg's $[h^{-1}]^a_{\phantom{a}b}(\theta)$ and this means that if $\theta^a(s)$ are the coordinates of $\gamma(s)$ the differential equation for $\exp (sX)$ is $$\begin{cases}\dfrac{d\theta^a}{ds}&=&[h^{-1}]^a_{\phantom{a}b}(\theta(s))X^b,\\ \theta^a(0)&=&0.\end{cases}$$
This shows to me that if Weinberg's path $\Theta^a_\theta(s)$ were some exponential path $\exp(sX)$ then the term $h^a_{\phantom{a}b}(\Theta_\theta(s))\frac{d\Theta_\theta^b}{ds}$ in the RHS of Eq. (4) would be just $X^b$. The issue is that Weinberg's path is arbitrary, so it need not satisfy that differential equation.

So in conclusion, it seems to me that what Weinberg is doing is taking the differential equation $$\gamma'(s)=[R_{\gamma(s)}]_{\ast e} X(s),\quad X(s) = [R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s)$$ valid for any curve, and translating it into the representation. Still I haven't fully formalized it yet.

 

[samalkhaiat]

Let \mathfrak{g}, G and M_{G} be the usual suspects. Objects on T_{e}(M_{G}) will be indexed by a,b,c, = 1,2, \ \cdots , \mbox{dim}(G) = n, and the coordinates, x of an arbitrary point in M_{G} will be labelled by \mu , \nu , \ … , = 1,2, \cdots ,n, but infinitesimal coordinates carry tangent space indices, \delta y^{a}.

Recall the group multiplication rule g(x)g(y) = g(z), where z^{\mu} = \varphi^{\mu} (x , y), is a set of real analytic functions, the structure functions, satisfying: \varphi^{\mu} \left( x , \varphi (y,z) \right) = \varphi^{\mu} \left( \varphi (x ,y) , z \right) ; \ \ \ \ \ \ \ \ \ (1)\varphi^{\mu} (x , 0) = \varphi^{\mu} (0 , x) = x^{\mu}; \ \varphi^{\mu}(x , \bar{x}) = \varphi^{\mu} (\bar{x} , x) = 0, \ \mbox{if} \ g^{-1}(x) = g(\bar{x}) .

Differentiating (1) with respect to z^{a}, we get \frac{\partial}{\partial z^{a}} \varphi^{\mu} \left( \varphi (x,y) , z \right) = \frac{\partial \varphi^{\mu}\left( x , \varphi (y,z)\right)}{\partial \varphi^{\rho}(y,z)} \frac{\partial \varphi^{\rho}(y,z)}{\partial z^{a}} .Now, we set z = 0 to obtain e^{\mu}{}_{a} ( \varphi ) = \frac{\partial \varphi^{\mu}(x,y)}{\partial y^{\rho}} \ e^{\rho}{}_{a}(y) , \ \ \ \ \ \ \ \ (2) where e(x) is a non-singular matrix defined by e^{\mu}{}_{a}(x) = \frac{\partial \varphi^{\mu}(x , y)}{\partial y^{a}}|_{y = 0} , \ \ \ e^{\mu}{}_{a}(0) = \delta^{\mu}_{a} . Introducing the inverse matrix e^{-1} = h: e^{\mu}{}_{a}(x) h^{a}{}_{\nu}(x) = \delta^{\mu}_{\nu}, Equation (2) become \frac{\partial}{\partial y^{\rho}} \varphi^{\mu} (x,y) = e^{\mu}{}_{a}( \varphi ) \ h^{a}{}_{\rho}(y) . \ \ \ \ \ \ \ ( \mbox{B} ) The (\mbox{B}) stands for Beauty. In fact, everything you ask about G and \mathfrak{g} will follow from Eq(B):

1) From G to \mathfrak{g}

From Eq(B), it follows that T_{a}(y) \equiv e^{\mu}{}_{a}(y) \frac{\partial}{\partial y^{\mu}} = e^{\nu}{}_{a}( \varphi ) \frac{\partial}{\partial \varphi^{\nu}} = T_{a}( \varphi ) , define a basis \big\{T_{a}\big\} of left-invariant vector fields in T(M_{G}), i.e., invariant under transformations corresponding to g(y) \to g(x) g(y) = g( \varphi ). Next, one uses the first integrability condition of Eq(B) to show that the corresponding vector space \mathfrak{g} = \big\{\theta^{a}T_{a}\big\} is closed under taking the Lie bracket between any two elements of \mathfrak{g} \big[ T_{a} , T_{b} \big] = C^{c}{}_{ab}T_{c}. And finally, the second (and the last) integrability condition of Eq(B) reproduces the Jacobi identities for the structure constants.

2) Riemannian metric and Haar measure

First, we write Eq(B) in the form h^{a}{}_{\mu}(\varphi ) \frac{\partial \varphi^{\mu}(x,y)}{\partial y^{\nu}} = h^{a}{}_{\nu}(y), or as \delta \varphi^{a}(x,y) = \delta y^{a} \equiv h^{a}{}_{\nu}(y) \mbox{d}y^{\nu} . This shows that the differentials \delta y^{a} are invariant with respect to left-multiplication by fixed group element, g(x) in our case, and using this result one can construct the left-invariant Haar measure and Riemannian metric d\mu (x) = \delta x^{1} \cdots \delta x^{n} = \mbox{Det}\left( h^{a}{}_{\mu}(x) \right) \mbox{d}x^{1} \cdots \mbox{d}x^{n} , ds^{2} = \eta_{ab} \delta x^{a} \delta x^{b} = g_{\mu\nu}(x) \mbox{d}x^{\mu} \mbox{d}x^{\nu}, where \eta_{ab} is any constant, positive metric and g_{\mu \nu}(x) = \eta_{ab} \ h^{a}{}_{\mu}(x) \ h^{b}{}_{\nu}(x) .

3) Setting y^{a} = \delta x^{a} in Eq(B) one finds \varphi^{\mu} ( x , \delta x ) = x^{\mu} + e^{\mu}{}_{a}(x) \ \delta x^{a} = x^{\mu} + \mbox{d}x^{\mu} . Thus,g(x + \mbox{d}x) = g(x) g(\delta x). This shows that right-multiplication by a group element close to the identity generates a small change in an arbitrary group element g(x). In particular, the derivative of a finite group element is \frac{\partial g (x)}{\partial x^{\mu}} = g(x) \ h^{a}{}_{\mu}(x) \left( \frac{\partial g(y)}{\partial y^{a}}\right)_{y = 0} . \ \ \ \ \ \ \ (\mbox{B}') Applying this to a representation g(x) \mapsto U(g(x)) , \ \ \ t_{a} = \left( \frac{\partial U(g)}{\partial y^{a}}\right)_{y = 0}, we find \frac{\partial U(g(x))}{\partial x^{\mu}} = U(g(x)) \ t_{a} h^{a}{}_{\mu}(x) , \ \ \ \ \ \ \ (\mbox{B}'') a result that is very useful in studying the Higgs mechanism.

4) From \mathfrak{g} to G

For any \theta^{a}T_{a} \in \mathfrak{g} there is a 1-parameter subgroup of G corresponding to a path in M_{G} whose tangent at the identity is \theta^{a}T_{a}. For s \in \mathbb{R} and coordinates x^{\mu} the path is defined by x^{\mu}(s), where \frac{\mbox{d}x^{\mu}(s)}{\mbox{d}s} = \theta^{a} \ e^{\mu}{}_{a}\left( x(s) \right), \ \ x^{\mu}(0) = 0 , \ \ \ \ (3) or \frac{\mbox{d}}{\mbox{d}s} g\left( x(s)\right) = \theta^{a}T_{a}\left( x(s)\right) \ g\left( x(s)\right) = \theta^{a} \ e^{\mu}{}_{a}\left( x(s)\right) \frac{\partial g}{\partial x^{\mu}} . \ \ \ \ (4)

To prove that this forms a subgroup, consider g\left( x(t) \right) g\left( x(s) \right) = g \left( \varphi \right), where \varphi^{\mu} = \varphi^{\mu} \left( x(t) , x(s)\right) . Using Eq(B) and Eq(3), we find \frac{\partial \varphi^{\mu}}{\partial s} = \frac{\mbox{d}x^{\nu}(s)}{\mbox{d}s} \frac{\partial \varphi^{\mu}}{\partial x^{\nu}(s)} = \theta^{a} \ e^{\mu}{}_{a} ( \varphi ) , \ \ \ \varphi^{\mu}|_{s = 0} = x^{\mu}(t) . This is identical to Eq(3), the solution is then given by \varphi^{\mu} \left( x(t) , x(s) \right) = x^{\mu}(t+s), implying that g\left( x(t) \right) g\left( x(s) \right) = g\left( x(t + s) \right). Thus \big\{ g \left(x(s)\right) \big\} forms an abelian subgroup of G depending on the real parameter s. At this point, we may define the map \exp : \mathfrak{g} \to G, blah, blah, blah using the B-C-H Formula, we can complete the construction of G from \mathfrak{g}.

5) Obtaining Equation (2.B.2) in Weinberg’s text

This is now a trivial task: Substituting Eq(B’) and Eq(3) in Eq(4), you find

\frac{\mbox{d}}{\mbox{d}s} g\left( x(s) \right) = \frac{\mbox{d}x^{\mu}(s)}{\mbox{d}s} \ g\left( x(s) \right) \ h^{a}{}_{\mu}\left( x(s) \right) \left( \frac{\partial g (y)}{\partial y^{a}}\right)_{y = 0} . Applying this to the representation g\left(x(s)\right) \mapsto U_{x}(s), we get \frac{\mbox{d}}{\mbox{d}s} U_{x}(s) = t_{a} U_{x}(s) \ h^{a}{}_{\mu} \left( x(s)\right) \frac{\mbox{d}x^{\mu}(s)}{\mbox{d}s} .