0
$\begingroup$

In Duistermaat's and Kolk's $\textit{Lie Groups}$ on page 2, they say that for a Lie group $G$ with $\mathfrak{g}=T_1G$ we can use the fact that $x\rightarrow xy$ and $y\rightarrow xy$ are identity maps if $y=1$ and $x=1$ respectively and they combine this with: $$\left (\begin{matrix} X \\ Y \end{matrix}\right ) = \left (\begin{matrix} X \\ 0 \end{matrix}\right ) + \left(\begin{matrix} 0 \\ Y \end{matrix}\right)$$ to obtain: $$T_{1,1}\mu:(X,Y)\rightarrow X+Y:\mathfrak{g}\times\mathfrak{g}\rightarrow \mathfrak{g}$$

$\textbf{Question:}$ How did this happen? I don't know what they are doing here. Why are they using those identity maps in the beginning and how does it relate to what seems to be written down as COLUMN VECTORS??? They don't even say what $X$ and $Y$ are although from context it seems to belong to $\mathfrak{g}$.

They go further on to say:

Applying this to $x \rightarrow xi(x)\equiv 1$ and using the sum rule for differentiation gives: $$T_1 i: X\rightarrow -X: \mathfrak{g}\rightarrow \mathfrak{g}$$ so these two mappings define an additive group structure on $\mathfrak{g}$ making $\mathfrak{g}$ itself a commutative Lie group.

$\textbf{Question:}$ How does the sum rule for differentiation come in at this last part? I don't see it.

P.S: This is the $\textbf{only}$ place where I've seen any mention of a Lie algebra being treated as a Lie group in itself in this manner. No other source seems to mention it at all.

$\endgroup$
4
  • $\begingroup$ This is the source of my confusion. They literally say that defining these two mappings makes $\mathfrak{g}$ into a commutative Lie group. No other source does this. $\endgroup$ Commented Sep 17, 2019 at 12:44
  • 1
    $\begingroup$ @Dietrich: The Lie algebra is a vector space. Under this structure, it is a commutative Lie group. Though I, too, have never seen anyone point this out like it matters. Calling it a commutative Lie group completely ignores the Lie bracket! $\endgroup$ Commented Sep 17, 2019 at 12:51
  • $\begingroup$ I won't have time to write up a full answer for awhile, but the idea is the consider the composition $G\rightarrow G\times G\rightarrow G$ where the second map is multiplication and the first map is either of the inclusions $g\mapsto (g,e)$ or $(e,g)$. This total composition is the identity map, hence induces the identity map $\mathfrak{g}\rightarrow \mathfrak{g}$, and then you can use this to understand the map $\mathfrak{g}\oplus\mathfrak{g}\rightarrow \mathbb{g}$ induced by multiplication. For the inverse map $i:G\rightarrow G$, consider the composition .... $\endgroup$ Commented Sep 17, 2019 at 12:54
  • $\begingroup$ $G\rightarrow G\times G\rightarrow G$ where the second map is, again, multiplication, and the first maps $g$ ti $(g,i(g))$. This total composition is constant, so has derivative $0$. But you now understand the map $\mathfrak{g}\oplus \mathfrak{g}\rightarrow \mathfrak{g}$, and this allows you to understand what $i$ does on the Lie algebra level. $\endgroup$ Commented Sep 17, 2019 at 12:55

1 Answer 1

1
$\begingroup$

A Lie algebra is a vector space (plus some more structure) and as such it also has a group structure via addition of vectors. Addition is also continuous so any (finite dimensional) vector space is a Lie group, just not a very exciting one.

What he seems to be doing in the beginning is showing that if you want to define a Lie group structure on a Lie algebra that is compatible with the Lie group, the trivial one is the natural choice to make.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .