# Determining the manifold picture of a Lie group — and thus determining global properties

1) How do we determine a Lie group's global properties when the manifold that it represents is not immediately obvious?

Allow me to give the definitions I am working with.

A Lie group G is a differentiable manifold G which is also a group, such that the group multiplication and the map sending $$g \in G$$ to its inverse $$g^−1 \in G$$, are differentiable $$(C^\infty)$$ maps.

As a Lie group is a manifold, one can parametrize the elements g in a small neighbor- hood of the identity e of G, by n real parameters $$\{x_1, . . . , x_n\}.$$ This is simply written as $$g = g(x_1, . . . , x_n)$$ and the parametrization is usually chosen in such a way that $$e = g(0, . . . , 0).$$ The number of independent real parameters (n) is called the dimension of the Lie group.

A Lie group G is connected iff $$\forall g_1, g_2 \in G$$, there exists a continuous curve connecting the two, i.e. there exists only one connected component.

A Lie group G is simply connected (if all closed curves on the manifold picture of G) can be contracted to a point.

A Lie group is compact if there are no elements infinitely far away fro the others.

The Lie group U(1) is quite easily identified as a circle in its manifold picture. This is connected, not simply connected, and compact.

However, SO(3) can (apparently) be viewed as manifold as such: a filled sphere of fixed radius, with antipodes identified. Once that has been realized, determining the global properties are straight forward. Getting there is another question, and I believe related to answering...

2) How do we algorithmically determine the parameter space of a Lie group – thus seeing it as a manifold?

For instance, for SU(2), we can write the matrix elements as complex, or decomposed with reals + i(reals), and use the det = 1 condition to determine that the Lie group is 3-deimensional. The step from 3-dimensional to a 3-sphere is not clear to me.

• I'm not really sure what you're asking for here. What data, exactly, is given, and what shall be derived? Oct 13, 2019 at 9:32
• @ACuriousMind I think OP is suggesting they're given a matrix lie group defined not by a parametrisation but by some algebraic properties. So similar to how with SO(3) one can obtain the Lie algebra and exponentiate it to find a parametrisation of the group. I'm unclear how much easier it is to find a parametrisation of the Lie Algebra in general though. Oct 13, 2019 at 11:27
• @jacob1729 or even just starting from the group! How do I use the conditions of the group, as I spoke of with SU(2), to determine what manifold the group is "dual" to? Oct 13, 2019 at 13:13

The underlying manifolds of the classical Lie groups are Stiefel manifolds $$V_k(\mathbb{F}^n)$$. These manifolds can be obtained as constraint surfaces in vector spaces over the real, complex and the quaternionic fields: $$\mathbb{F} = \mathbb{R} ,\mathbb{C} \, \mathrm{or}, \mathbb{H}$$. The constraint equations is given by the set of equations (From the Wikipedia article). $$V_k(\mathbb{F}^N) = \{ A\in \mathbb{F}^{n\times k} | A^*A = I_{k\times k} \}$$
{\begin{aligned}V_{n}(\mathbb {R} ^{n})&\cong \mathrm {O} (n)\\V_{n}(\mathbb {C} ^{n})&\cong \mathrm {U} (n)\\V_{n}(\mathbb {H} ^{n})&\cong \mathrm {Sp} (n)\end{aligned}}}
{\begin{aligned}V_{n-1}(\mathbb {R} ^{n})&\cong \mathrm {SO} (n)\\V_{n-1}(\mathbb {C} ^{n})&\cong \mathrm {SU} (n)\end{aligned}}}
One practical use of this constrained realization is in the integration theory over Haar measures; where the integration over the constraint surfaces is usually easier than the integration other parametrizations such as Euler angles (In the case of $$SU(2)$$ it is just an integration over the round three sphere).