$\mathbb{C}$ as the Lie algebra of $\mathbb{C}$

Question

It is known that $\mathbb{C}$ is a linear algebraic group as the set of matrices $\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}$ with $a \in \mathbb{C}$. I read in some notes that $Lie(\mathbb{C})$, the Lie algebra associated to $\mathbb{C}$, is isomorphic to $\mathbb{C}$, but I cannot see why. I only know that if $G$ is a linear algebraic group, $Lie(G)$ is the set of all $\phi : O(G) \to O(G)$ (where $O(G)$ is the set of regular functions on $G$) that satisfy Leibniz rule and such that $g \cdot \phi(f) = \phi(g \cdot f), \forall g \in G, \forall f \in O(G)$. It looks hard to use this definition, so I was thinking of using the fact that $Lie(\mathbb{C})$ is isomorphic to the tangent space, but I am not really sure how. Any hints?

Answer 1

Actually, if you want to see $\Bbb C$ as a linear group, you should see it isomophic to$$\left\{\begin{bmatrix}1&a\\0&1\end{bmatrix}\,\middle|\,a\in\Bbb C\right\};$$that set that you mentioned is not a group.

Clearly, this group is $1$-dimensional. Therefore, its Lie algebra $\mathfrak g$ is a $1$-dimensional complex Lie algebra. And the only way of defining a Lie braket $[\cdot,\cdot]$ on a $1$-dimensional vector space $V$ is to define $[x,y]=0$ for every $x$ and every $y$ in $V$. So, yes, it is naural to see $\mathfrak g$ as $\Bbb C$ endowed with the trivial braket.

Answer 2

[In your setting both the group $G$ and the underlying field $k$ "are" the complex numbers (once, its additive group viewed as algebraic group, once as field), which can be confusing. So in the following I will use the notations $k$ and $G$. If you insert "$\mathbb C$" for both, see what a mess this would become!]

If you want to use the language in your question, you can also proceed by showing:

• $O(G)$ is isomorphic to the polynomial ring $k[T]$ where further
• the element $a \in G$ acts on a polynomial $p(T)$ via $p(T) \mapsto p(T+a)$

Then an element of the Lie algebra as defined in your post is a $k$-linear map $\phi: k[T] \rightarrow k[T]$ such that $\phi(f\cdot g) = \phi(f) \cdot g + f \cdot \phi(g)$. Show that this condition entails that $\phi$ is completely determined by $\phi(T)$, and that $\phi(c)=0$ for all constants $c$. Then, the compatibility with the $G$-action says that, if $\phi(T) = \sum c_iT^i$, we have

$$ \phi(T+a)= \sum c_i (T+a)^i$$

for all $a \in G$, but by $\phi$ being linear and our already established $\phi(a)=0$, the left hand side is just $\phi(T) = \sum c_i T^i$. Show that this is impossible unless $c_i = 0$ for all $i \ge 1$. Consequently $\phi(T) = c_0$ is a constant, and you can now show that $\phi$ must have been the differential operator $c_0 \cdot \dfrac{d}{dT}$ all along.

So $Lie(G)$, as a $k$-vector space, is the set $\{c_0 \cdot \dfrac{d}{dT} : c_0 \in k\}$ which obviously is of $k$-dimension $1$. You can now either use the fact that in any $1$-dimensional Lie algebra every Lie bracket is $=0$, or use whatever definition of the Lie bracket is given in your source to see that it is identically $0$ here. Probably all you need for that is that $G$ is abelian.

Worthwhile: Look at the two alternative definitions of the Lie algebra of an algebraic group in chapter 12b of Milne's online book draft, and see how they identify with our result above. (Most directly you should be able to identify, in 12.5, our $\phi$ with Milne's $D$ once you see that in our setting, $I/I^2 = T \cdot k[T]/ T^2 \cdot k[T]$.) Note also that then your example is a special case of both 12.6. and 12.7.