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In physics, we want irreducible representations of the symmetry group of our system. However, one frequently sees representations of the corresponding Lie algebra being studied instead. Is it that in the specific groups that pop up in physics for some reason there is a bijection between irreps of the group and irreps of its Lie algebra? I know that one can get representations of the Lie algebra by differentiating the representations of the Lie group, but I also know that not all representations of the Lie algebra can be obtained in this way.

A guess: is it because we don't really want representations but instead we want projective representations, and these are in a bijective relation with the representations of the Lie algebra (or some extension)?

I find the literature on this pretty unclear.

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  • $\begingroup$ Reviewed rotations? $\endgroup$ Commented Jul 22, 2019 at 10:14

4 Answers 4

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The objects which matter in physics are Lie groups and not Lie algebras. Lie algebras approximate only infinitesimal group transformations and in quantum mechanics the finite and global properties of the transformations matter.

However, (considering quantum systems with a finite number of degrees of freedom), the spaces of quantum states are projective, as there is no physical meaning to the overall magnitudes and global phases of state vectors. Thus, symmetry groups act on the spaces of states via projective representations.

For a semisimple compact Lie group, a projective representation is a true representation of its (simply connected) universal covering. The representations of the universal covering group are in a $1-1$ correspondence with the representations of its Lie algebra (which is the same Lie algebra as the original group) . This is the reason why all representations of the group's Lie algebra can appear as realizations of symmetries in quantum systems.

May be the most famous case is the rotation group $SO(3)$, which can be parametrized by The Euler's angles. The true representations of the rotation groups are the integer spin representations. However, there are quantum systems in which the rotation symmetry is realized by means of the half-integer spin representation (such as the electron spin or a qubit). The half integer representations are only projective representations of the rotation group; however, they are true representations of its universal covering $SU(2)$. The representations of $SU(2)$ are in a $1-1$ correspondence to the representations of the isomorphic Lie algebras of both groups $\mathfrak{so}(3) \cong \mathfrak{su}(2)$.

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    $\begingroup$ And how do central extensions "join the party"? $\endgroup$
    – soap
    Commented Jul 23, 2019 at 9:39
  • $\begingroup$ What about the commutator in Quantum Mechanics (which is zero in classical mechanics) for which one needs the group operation but also a 'minus' type operation ? $\endgroup$ Commented Jul 23, 2019 at 11:04
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    $\begingroup$ @Soap The answer treats only quantum mechanical systems with a finite number of degrees of freedom having a compact semisimple Lie group of symmetries. Sorry for not writing that explicitly. The central extentions of semisimple Lie groups are just the covering groups. When the group is not semisimple, then a projective representation is a true representation of a central extension of the group, which can be non-discrete in this case. $\endgroup$ Commented Jul 23, 2019 at 11:29
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    $\begingroup$ @Andre Holzner The paralell object for Lie groups is the commutant. Given two group elements $g$ and $h$, their commutant is $ghg^{-1}h^{-1}$. Furthermore, the whole Cartan-Weyl theory (roots, weights etc.) can be developed on the basis of Lie groups only without Lie algebras, please see chapter 2 in Pressley and Segal loop groups book books.google.co.il/… $\endgroup$ Commented Jul 23, 2019 at 11:41
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    $\begingroup$ @Soap The general statement is a consequence of Wigner's theorem, please see, for example, the folowing lecture notes by Pflaum (page 19)libermath.org/GeometryClassicalAndQuantumFields/pdf/…. The special case of semisimple groups stems from Whitehead Lemmas which asserts that semisimple Lie algebras have no nontrivial cental extensions, therefore the central extensions of the corresponding groups must be discrete, please see for example Fiallos work math.ubc.ca/~reichst/Lie-Algebra-Cohomology.pdf $\endgroup$ Commented Jul 23, 2019 at 16:36
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  1. It a general fact that any Lie group representation induces a corresponding Lie algebra representation (but not necessarily the other way around). Therefore, short of topological informations, we can often learn a great deal about physics by dealing with the Lie algebra (which in practice is a mathematically easier gadget to handle, namely just a vector space).

  2. Of course, a full treatment would include investigating whether the Lie algebra representations can be lifted to consistent (possible projective$^1$) Lie group representations of the theory. This analysis is often glossed over in physics textbooks.

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$^1$ Whether projective representations are allowed depends on context.

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  • $\begingroup$ But, as far as I understand, one is always interested in projective representations of the group, and the way they connect with the ordinary representations of the Lie algebra is unclear to me. For example, when dealing with the rotation group we take its universal cover, and its representations are in a one-to-one correspondence with the projective representations of the rotation group; in the other hand, when dealing with projective representations of the loop group we take the central extension instead (not the universal cover). $\endgroup$
    – soap
    Commented Jul 22, 2019 at 11:18
  • $\begingroup$ So: my problem is specifically with how exactly one uses ordinary representations of Lie algebras to get the projective representations of the groups. $\endgroup$
    – soap
    Commented Jul 22, 2019 at 11:23
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    $\begingroup$ @Soap We don't always use projective reps. For example, matter fields come as tensor representations of the gauge group, not projective ones. $\endgroup$ Commented Jul 22, 2019 at 22:13
  • $\begingroup$ @Soap Ordinary reps of the Lie algebra gives in some cases projective reps of the group, because the Lie algebra does not differentiate between a group and its double cover. An example is SO(3) and SU(2). $\endgroup$
    – timur
    Commented Nov 13, 2020 at 2:54
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Observables often close on a Lie algebra, and their matrix elements are directly related to measurable quantities, v.g. average values, eigenvalues (spectrum), transition rates to name a few.

In addition, there are some very useful algebras for which there is no group, v.g. the Temperley-Lieb algrebra, $q$-deformations, etc.

Finally, the equations of physics tend to be expressed in differential form, comparing functions that are infinitesimally close. Thus it’s no surprise that infinitesimal generators, i.e. elements of the algebra, tend to be more useful than group elements, which require (sometimes very intricate) exponentiation.

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  • $\begingroup$ Can you please explain what do you mean when you say that the Temperley–Lieb algebra has no group? Isn't an algebra an Abelian group with respect to one operation (additon) by definition? $\endgroup$ Commented Jan 26, 2023 at 18:17
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It's not true that we only care about groups. Instead, for most applications Lie algebras are actually more important.

To quote Ed Witten in his review on "Physics and Geometry"

Experiment tells us more directly about the Lie algebra of G than about G itself. When I say that G contains the subgroup SU(3) X SU(2) x U(1), I really mean only that the Lie algebra of G contains that of $SU(3) \times SU(2) \times U(1)$; there is no claim about the global form of $G$. For the same reason, in later comments I will not be very precise in distinguishing different groups that have the same Lie algebra.

Or to quote Sidney Coleman

In high-energy theory. we tend to focus on the Lie algebra of a group and ignore its global structure;

Moreover, the following quote from a recent paper by David Tong may be helpful

We learn in kindergarten that we should take $$ \tilde{G} = U(1)\times SU(2) \times SU(3) $$ But this is not quite accurate. Experimental considerations tell us only that the gauge group is $$ G = {\tilde{G}}/{\Gamma} $$ where $\Gamma$ is a discrete group. At present, we can only say that the gauge group involves a quotient by $\Gamma$, which is a subgroup of ${\bf Z}_6$, i.e. $$ \Gamma = {\bf Z}_6,\ {\bf Z}_3,\ {\bf Z}_2\ {\rm or}\ {\bf 1} $$ Each of these possibilities defines a different theory and, ultimately, gives rise to different physics. The obvious questions are: which describes our world? And how can we tell? [...] Correlation functions of local operators in ${\bf R}^{1,3}$ depend only on the Lie algebra of the gauge group and are unaffected by global issues such as the choice of $\Gamma$. This means that no current experiment can distinguish between the four possibilities. Nonetheless, the physics in flat space can depend in subtle ways on $\Gamma$ (and in more dramatic ways when spacetime has interesting topology). The purpose of this paper is to describe the crudest differences between the theories: the spectrum of line operators and the periodicities of theta angles.

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