I am a bit confused because I think I have a proof for a statement which I think is wrong.
The statement is:
\begin{equation}
\text{Every compact simply connected Lie group has even dimension.}
\end{equation}
However, if I remember correctly $S^3$ has a Lie group structure.
I will now elaborate on my "proof" for the above statement.
Corollary 21.6 in Milnor´s Morse Theory says the following:
A simply connected Lie group $G$ with bi-invariant (Riemannnian) metric splits as a cartesian product $G'\times \mathbb{R}^k$, where $G'$ is compact and the Lie algebra of $G'$ has trivial center.
Let now $G$ be a compact simply connected Lie group. As $G$ is compact it posseses a bi-invariant Riemannian metric $\langle \cdot,\cdot \rangle$. By the corollary, the Lie algebra $\operatorname{Lie}(G)$ of $G$ has trivial center, i.e. for every non-zero $X\in \operatorname{Lie}(G)$ there exists $Y \in Lie(G)$ s.t. $[X,Y]\neq 0$.
Define a bilinear form $q$ on $Lie(G)$ by $q(X,Y)=[X,Y]$. As this is bilinear, there ex. a linear transformation
\begin{equation}
A: \operatorname{Lie}(G)\to \operatorname{Lie}(G)
\end{equation}
such that
\begin{equation}
q(X,Y)=\langle AX,Y\rangle \quad \text{for all } X,Y\in \operatorname{Lie}(G).
\end{equation}
Since $\operatorname{Lie}(G)$ has trivial center $A$ is injective and hence an automorphism. The fact that $q$ is skew-symmetric translates into
\begin{equation}
A^T=-A.
\end{equation}
If we denote the dimension of $G$ by $n$, this implies
\begin{equation}
\det(A)=\det(A^T)=(-1)^n\det(A).
\end{equation}
However, as $A$ is an automorphism we know $\det(A)\neq 0$. Therefore, $n$ is even.
I would be grateful for clarifications. That can either be pointing out a mistake in my "proof" or any other mistake I made.
Thanks in advance.
Best Answer
A bilinear form is supposed to be real-valued. Your $q$ is $\operatorname{Lie}(G)$-valued. So the theorem you're appealing to in order to get your linear transformation $A$ doesn't apply.