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I would like to clarify something and I will take the example of the $SO(2)$ Lie group.

A Lie group is a manifold with a group structure. Thus we can define maps on it to be able to move on the manifold.

In $SO(2)$, I can write the matrices as :

$$ \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} $$

As I understand things, I see the $\theta$ as a map on the manifold.

But when we want to find the Lie algebra of the group, we say that it is the tangent space to the identity.

Thus, we have to take a curve from $\mathbb{R}$ to $SO(2)$ that pass in the identity and we have to derive it on the identity to find the Lie algebra.

In this vision, we could say that $\theta$ is in fact the parametrisation of my curve.

Thus an element of the Lie algebra is the derivative of the matrix according to $\theta$, and I find :

$$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$

Finally : is $\theta$ a map on my manifold or is it a curve on it ?

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    $\begingroup$ We have $SO(2) \cong S^1$ so $\theta$ would be an argument. This is not very well defined as $\theta = \theta + 2\pi$. What you found is a surjection from $\Bbb R \to S^1$, $t \mapsto e^{it}$. If you restrict it this is also a curve passing by zero. Also notice that $ \mathfrak{so(2)} \cong \Bbb R$ since $S^1$ is one-dimensional, so this is not a very interesting Lie algebra. $\endgroup$ – user171326 May 16 '17 at 13:21
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The map $\epsilon : \Bbb R \to SO(2)$ defined by the first display equation, namely, $$\epsilon : \theta \mapsto \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta}$$ is a perfectly good parameterized curve on $SO(2)$. (In fact, it is a homomorphism of Lie groups, since $\epsilon(\alpha + \beta) = \epsilon(\alpha) \epsilon(\beta)$.)

This map is surjective, so it defines a parameterization of $SO(2)$. On the other hand, the map is not injective, so it is not bijective, that is, without making an additional choice, we cannot immediately regard $\theta$ as a map on (some subset of) $SO(2)$.

We can, however, choose an open interval $I \subset \Bbb R$ on which $\epsilon$ is injective, in which case $\epsilon\vert_I^{-1}$ is a homeomorphism $\epsilon(I) \to I$; in particular, this inverse is a smooth local chart on $SO(2)$ and hence defines a (preferred) coordinate on $\epsilon(I) \subset SO(2)$.

So, we may as well (somewhat abusively) call this coordinate $\theta$, or equivalently, just use $\theta$ to denote the map $\epsilon_I^{-1}$. In this sense, $\theta$ is a map (in particular satisfying $\epsilon \circ \theta = \textrm{id}_{\epsilon(I)}$), but this declaration depends on our (noncanonical) choice of $I$. Note, by the way, that there is no choice of $I$ that produces a global chart, that is, for which $\epsilon(I) = SO(2)$.

As N.H. hints in his helpful comment, there is another way to view the charts $\varepsilon\vert_I^{-1}$: We can identify $SO(2)$ with $\Bbb S^1 \subset \Bbb C$ via the identification $$\varepsilon(\theta) \leftrightarrow \exp (i\theta) .$$ Then, unwinding definitions shows that $\varepsilon\vert_I^{-1}$ coincides via this identification with the restriction $\arg\!\vert_{\epsilon(I)}$ of some choice of branch of the argument function.

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  • $\begingroup$ Thank you. To summarize : a parametrisation is not necesseraly bijective but it is surjective (it must go on all the manifold). A curve is just a map from $I \subset \mathbb{R}$ in the manifold. A map is a map form $U \subset R^n$ to the manifold and it is a diffeomorphism so it must be bijective. In my example, if I restrict $\theta$ to $[0; 2 \pi[$, my function $f(\theta)$ will be at the same time a curve and a map of my manifold. But if I don't put any restriction on $\theta$, it will just be a curve as it will not be bijective. $\endgroup$ – StarBucK May 16 '17 at 13:45
  • $\begingroup$ A parametrisation will always be a curve if I work on an 1 dimension manifold. Do you agree with all that I said ? $\endgroup$ – StarBucK May 16 '17 at 13:46
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    $\begingroup$ You're welcome. Note that the requirement that a parameterization be surjective is not universal. Sometimes people will refer to nonglobal parameterizations as local parameterizations, sometimes just as parameterizations. For me, maps are always continuous, and certainly curves should be. The sentence "A map is a map from..." is both circular and confusing. I would refer to a homeomorphism from an open set $U \subset \Bbb R^n$ to an open subset of manifold also as a parameterization of the manifold, and its inverse as a coordinate chart. $\endgroup$ – Travis May 16 '17 at 13:51
  • $\begingroup$ Also, when working with these concepts, often it is best to restrict to working with open intervals and open sets more generally. One can deal with more general sets, but this often involves technicalities that are a consequence of the fact that boundary points behave qualitatively differently than interior points. $\endgroup$ – Travis May 16 '17 at 13:53
  • $\begingroup$ Finally, as both of the answers point out, the Lie group $SO(2)$ is highly nonrepresentative here, as it is itself $1$-dimensional, and (being connected) is hence parameterizable by regular curve. So it would be useful to supplement this example with one involving a higher-dimensional Lie group. $\endgroup$ – Travis May 16 '17 at 13:55
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First, I would like to emphasize that a Lie group is a smooth manifold equipped with a group structure whose operations (multiplication, inverse) are smooth. It is not enough to ensure that $G$ is a manifold and a group to deduce that $G$ is a Lie group.

Regarding your question, $\theta$ is not a map. However, the following map is a parametrization of $\textrm{SO}(2)$ : $$\theta\mapsto\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}.$$ As you deduced it, $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is in $\mathfrak{so}(2)$. Furthermore, according to the above parametrization of $\textrm{SO}(2)$, the Lie algebra $\mathfrak{so}(2)$ has dimension $1$ over $\mathbb{R}$. Whence, $\mathfrak{so}(2)$ is the set of $2\times 2$ skew-symmetric matrices.

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First of all: it is incorrect to say that $\theta$ is a map/parameterization/curve. The map is the function $$ f(\theta) = \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta} $$ It is $f$ which is the map, not $\theta$. $\theta$ is just a varaible, which is an argument of the function $f$. If we consider this function over the domain $[0, 2 \pi]$, then we have a parameterization of $SO(2)$.

Second: the Lie algebra is the vector space of all derivatives of curves through the identity. Note that $$ \frac d{d \theta}f( \alpha \theta) = \alpha \pmatrix{0 & -1 \\1 & 0} $$ and we see that the Lie algebra is actually a $1$-dimensional space. Note that there are infinitely many paths through the identity, corresponding to varying speeds and directions of traversal.

$SO(2)$ happens to be a connected one-dimensional Lie group. As such, your parameterization is both a curve on the Lie group and a parameterization of the entire thing.

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