2
$\begingroup$

I apologize in advance if my question is elementary. Before I present my question I mention my motivation:

Motivation:

A Lie group is a manifold. At the same time it is a Riemannian manifold equipped with an invariant metric. Moreover its tangent space at the neutral element is a Lie algebra.

On the other hand every manifold can be embedded in some Euclidean space. Furthermore this embedding can be chosen to be an isometric embedding. Moreover every lie algebra can be embedded in an inner Lie algebra, an algebra with an inner Lie bracket $[a,b]=ab-ba$.

In our question, for a given Lie group, we would like to combine all (or parts of) these properties.

Question:

Let $G$ be a Lie group with neutral element $e$. Is there an algebra structure on some $\mathbb{R}^n$ and a smooth embedding $f:G \to \mathbb{R}^n$ with $f(e)=0$ such that $Df_e$ carries $T_e G$ to a sub Lie algebra of $\mathbb{R}^n$ with the inner Lie structure $[a,b]=ab-ba$?Can we choose such $f:G \to \mathbb{R}^n$ an isometric embedding where $G$ is equipped with an invariant metric and $\mathbb{R}^n$ with its standard metric? Can we choose such an $f$ such that $Df_g$ carry all tangent spaces $T_g G,\;g\in G $ to a Lie Subalgebra of $\mathbb{R}^n$ as a Lie algebra embedding?

A trivial example:

The embedding of circle in the plane, when we equipe the plane with an arbitrary commutative algebra structure, satisfies all requirements of the question.

$\endgroup$
  • 2
    $\begingroup$ Every (real) Lie group is a finite cover of a linear Lie group (i.e., a closed subgroup of some $\mathrm{GL}_m(\mathbb R)$), and this can obviously be done for linear Lie groups; so maybe it's enough to show that this condition is preserved under covers? (Or, if it's false, then a non-linear group like a spin or metaplectic group is the place to look.) $\endgroup$ – LSpice Feb 17 '18 at 22:50
  • $\begingroup$ @LSpice It is interesting to think about non linear ones. BTW I was not aware of the fact that every Lie group is a cover of a linear group. Thanks for this point and for your comment. $\endgroup$ – Ali Taghavi Feb 17 '18 at 23:00
  • 3
    $\begingroup$ @LSpice no (1) the universal covering of $SL_2(R)$ is not finite cover of a linear Lie group (2) the quotient of the Heisenberg group by an infinite cyclic central subgroup has no cover at all that is a linear Lie group. Also there are two definitions of linear: has an injective continuous representations, vs has one with closed image. These are equivalent, but it's a theorem (maybe of Mostow, I don't remember right now). $\endgroup$ – YCor Feb 17 '18 at 23:23
  • 1
    $\begingroup$ @AliTaghavi unfortunately there are two non-equivalent definitions of isometric embedding: isometric for the distance, or infinitesimally isometric (i.e. the pull-back of the Riemannian metric of the target space equals the Riemannian metric of the left-hand space). Which one do you have in mind (one can even imagine intermediate notions, such as "locally isometric", i.e. preserving small distances). $\endgroup$ – YCor Feb 17 '18 at 23:25
  • 1
    $\begingroup$ @AliTaghavi by algebra do you mean associative algebra? (there are interesting structures, called pre-Lie algebras or left-symmetric algebras, whose brackets are Lie brackets). Also, I guess that by "invariant metric" you mean "left-invariant metric"? $\endgroup$ – YCor Feb 17 '18 at 23:27
1
$\begingroup$

For a linear group, it would seem that the defining embedding $G\hookrightarrow GL(n, \mathbb{R}) \to M^{n\times n}$ works, no?

$\endgroup$
  • $\begingroup$ Thanks for your answer. yes but it is not an isometric embedding. $\endgroup$ – Ali Taghavi Feb 17 '18 at 22:56
  • $\begingroup$ Is there an isometric embedding of $GL(n,\mathbb{R})$ into some $M^{k\times k}$ whose derivative preserves the Lie algebra structure? $\endgroup$ – Ali Taghavi Feb 19 '18 at 8:11
  • $\begingroup$ @AliTaghavi, $\mathrm{GL}(n, \mathbb R) \to \mathfrak{gl}(n, \mathbb R)$ does it; its derivative is the identity map. $\endgroup$ – LSpice Feb 20 '18 at 2:49
  • $\begingroup$ @LSpice but it is not an isometry since the metric of $GL(n,R)$ is not the Euclidean metric. $\endgroup$ – Ali Taghavi Feb 20 '18 at 7:53
  • $\begingroup$ @AliTaghavi, sorry; I missed the word 'isometric'. $\endgroup$ – LSpice Feb 20 '18 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.