advanced34 min studyLesson 495 in Mathematics

Lie subgroups and Lie correspondence

Integrate subalgebras and homomorphisms while tracking immersed topology, covering groups, and period obstructions.

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What you’ll understand

  • State the subalgebra-to-connected-immersed-subgroup correspondence.
  • Use simple connectedness in the integration of a Lie algebra homomorphism.
  • Identify a period obstruction to descending a homomorphism to a quotient group.

The Lie algebra captures local multiplication, but topology determines how those local pieces close up globally. A subalgebra integrates to a connected immersed subgroup, which need not be closed in the ambient group. A finite-dimensional Lie algebra integrates to a connected simply connected group, while other connected integrations can differ by discrete central quotients. The same distinction appears for homomorphisms: a valid infinitesimal map can fail to respect the periods of a chosen global source group. We will keep these three correspondence statements separate.

A subalgebra produces an immersed subgroup

For a Lie subalgebra h of the Lie algebra of G, left translations of h form an involutive distribution. The Frobenius theorem integrates the distribution through the identity, producing a unique connected immersed Lie subgroup with that tangent algebra. The subgroup generated by exp h describes the same object. Its intrinsic manifold topology is part of the construction; it is not automatically the topology inherited as a subset of G.

An irrational line in the Lie algebra of a two-dimensional torus illustrates the issue. Exponentiating it gives an injectively immersed real line with dense image in the torus. It is not an embedded copy of the real line with the torus subset topology and is not a closed subgroup. By contrast, a rational-slope line closes into a circle. The tangent dimension is one in both cases, so tangent data alone cannot decide whether the image is closed.

h⊆g a Lie subalgebra⟷H⊆G a connected immersed Lie subgroup\mathfrak h\subseteq\mathfrak g\ \text{a Lie subalgebra}\quad\longleftrightarrow\quad H\subseteq G\ \text{a connected immersed Lie subgroup}
The correspondence is with connected immersed subgroups equipped with their Lie subgroup structure. Closed subgroups are embedded, but closedness is not asserted by this theorem.↔ Scroll wide equations horizontally.

Integrate the algebra before imposing global periods

Every finite-dimensional real Lie algebra is the Lie algebra of a connected simply connected real Lie group, unique up to an isomorphism inducing the specified algebra identification. For connected simply connected G, every Lie algebra homomorphism from its algebra to the algebra of another Lie group H integrates uniquely to a smooth group homomorphism G→H. The target does not need to be simply connected. The source assumption removes ambiguity caused by continuing a local map around nontrivial loops.

For a connected source that is not simply connected, first integrate on its universal covering group. The map descends to the original source exactly when it kills the covering kernel. That kernel is discrete and central, and its image is not determined merely by checking the infinitesimal bracket equation. For example, all scalar linear maps between one-dimensional abelian Lie algebras preserve brackets, but only particular scalars respect the circle's integer periods.

G connected and simply connected,φ:g→h a Lie algebra homomorphism⟹∃! Φ:G→H, dΦe=φG\ \text{connected and simply connected},\quad \varphi:\mathfrak g\to\mathfrak h\ \text{a Lie algebra homomorphism}\quad\Longrightarrow\quad\exists!\,\Phi:G\to H,\ d\Phi_e=\varphi
For a non-simply-connected connected source, the integrated map on its universal cover must annihilate the discrete covering kernel before it descends.↔ Scroll wide equations horizontally.

Local isomorphism is weaker than global isomorphism

The additive real line and the circle both have a one-dimensional abelian Lie algebra, but their topology differs. More generally, a connected Lie group is a quotient of its simply connected covering group by a discrete central subgroup. This preserves the Lie algebra while changing global periods, fundamental groups, and which representations descend. It does not say every abstract central subgroup is an admissible quotient: discreteness supplies the appropriate covering behavior in this description.

When comparing two integrations, list the simply connected group, the chosen discrete kernel, and the proposed map's action on that kernel. For a subgroup inside an existing ambient group, separately ask whether its immersed image is closed. These questions are related through topology but are not interchangeable. A dense immersed subgroup can be the correct analytic subgroup for a subalgebra while failing to provide a Hausdorff homogeneous-space quotient with the ordinary subset quotient topology.

A bracket-preserving scalar need not descend to the circle

Identify the circle as t↦eitt\mapsto e^{it} with period 2π2\pi. Which real scalar maps t↦ct on its Lie algebra integrate to homomorphisms S¹→S¹? Check c=3 and c=1/2.

  1. On the simply connected source R, every real c gives the smooth homomorphism t↦eictt\mapsto e^{ict}. The Lie algebra is abelian, so multiplying by c always preserves its zero bracket.

  2. To descend through R/(2πZ), parameters t and t+2π must have the same image. Their image ratio is e2πice^{2\pi i c}, which equals one exactly when c is an integer. This is the covering-kernel condition in a concrete form.

  3. For c=3 the descended map is z↦z³ and its kernel consists of the three cube roots of one. For c=1/2, t=0 and t=2π describe the same source point but map to one and minus one, so there is no well-defined descended homomorphism.

The result

The infinitesimal equation admits every real scalar; the chosen circle topology permits only integer winding numbers.

Explore the model

Interactive lab

Interactive lab · modular

Inspect a finite period before comparing continuous ones

Pause, Reset, then set modulus m=12 and additive step=5. Predict orbit period12. Repeat from Reset with both m=12 and step=4; predict period3, and Step through the residues. This finite additive orbit illustrates a period constraint but is not a continuous torus flow, an irrational-density simulation, or a test of simple connectedness.

Inspect a finite period before comparing continuous onesAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA local increment and a global period are different inputs; changing the increment can change which points are identified along an orbit.

Check your understanding

A Lie algebra homomorphism is given from the algebra of a connected simply connected real Lie group G to that of H. Which conclusion holds?

Try it yourself

For t↦(eit,e2it/3)t\mapsto(e^{it},e^{2it/3}) in the two-torus, find the smallest positive period and describe the connected subgroup. Contrast only the period conclusion with replacing 2/3 by √2.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The first coordinate requires t=2πk. Substitute that into the second-coordinate period condition.

Compare with the solution

The second coordinate then requires 2k/3 to be an integer. The smallest positive k is three, giving period6π. The image is an embedded closed circle parametrized by R/(6πZ). With slope √2, a nonzero period would require both k and k√2 to be integers, impossible. The resulting one-parameter map is injective; the additional standard irrational-torus-flow theorem says its image is dense, a claim stronger than the period calculation alone.

Take it with you

  • Subalgebras correspond to connected immersed subgroups, not automatically closed ones.
  • A finite-dimensional algebra has a unique connected simply connected integration up to the specified isomorphism.
  • Descending a map from a universal cover requires respecting its discrete central kernel.

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