Different definitions for semisimple Lie group

For simplicity, all Lie groups below are connected.

My preferred definition is that a Lie group $G$ is semisimple if its Lie algebra $\mathfrak{g}$ is semisimple. This is equivalent to definition 1 since solvable ideals in $\mathfrak{g}$ exponentiate to connected solvable normal subgroups of $G$ and vice versa.

Proposition: Definition 2 implies definition 1.

Sketch. Recall that one definition of a reductive Lie algebra $\mathfrak{g}$ is that it is a Lie algebra whose adjoint representation is semisimple. Any Lie algebra of real or complex matrices closed under conjugate transpose $X \mapsto X^{\dagger}$ (this is equivalent to $G$ being closed under conjugate transpose) has this property because the form

$$X, Y \mapsto \text{Re}(\text{tr}(X^{\dagger} Y))$$

is a positive-definite (and in particular nondegenerate) invariant symmetric bilinear form on such matrices, so we can take the complement of a subrepresentation of the adjoint representation and get another subrepresentation.

A reductive Lie algebra is the direct sum of an abelian subalgebra (their center) and a semisimple algebra (their commutator), and hence is semisimple iff it has trivial center, which is equivalent to $G$ having discrete center. $\Box$

In general, definition 1 does not imply definition 2. For example, the universal cover $\widetilde{SL}_2(\mathbb{R})$ of $SL_2(\mathbb{R})$ is semisimple in the sense that its Lie algebra $\mathfrak{sl}_2(\mathbb{R})$ is semisimple, but its center is $\mathbb{Z}$.

However, one can say the following. If $\mathfrak{g}$ is a semisimple Lie algebra, then it admits a Cartan involution, namely an involution $\theta : \mathfrak{g} \to \mathfrak{g}$ such that

$$B_{\theta}(X, Y) = B(X, \theta Y)$$

is negative definite, where $B$ is the Killing form. If $\text{ad}_X : \mathfrak{g} \to \mathfrak{g}$ denotes the adjoint action of $X \in \mathfrak{g}$, then the adjoint (!) of $\text{ad}_X$ with respect to $B_{\theta}$ turns out to be $-\text{ad}_{\theta X}$ by direct calculation, so the adjoint representation exhibits $\mathfrak{g}$ as a Lie algebra of real matrices closed under transpose, and hence exhibits $G/Z(G)$ as a Lie group of real matrices closed under transpose. So definition 1 implies definition 2 up to taking covers, or equivalently up to discrete center. I don't know off the top of my head if requiring $G$ to be a matrix group fixes this.