Is the condition ``adjoint action does not have eigenvalue $-1$" dense in a Lie group?

Question

I need to answer (affirmatively, I hope) the following question:

In a Lie group $G$ whose Lie algebra $\mathfrak{g}$ is equipped with an $\mathrm{Ad}$-invariant scalar product, is the open subset $$\{g \in G \mid \operatorname{ker} (1 + \operatorname{Ad}_{g}) = \{0\}\}$$ dense in $G$?

In the case $G=\mathrm{GL}_n\mathbb{C}$, an affirmative answer goes as follows, which works for any complex reductive group as well. First we can restrict to the open dense subset $U$ of matrices with distinct eigenvalues. By conjugating any $g\in U$ to a diagonal matrix, we can see that $\mathrm{Ad}_g$ does not have eigenvalue $-1$ if and only if the proportion of any two eigenvalues is not $-1$, and this condition is dense.

However, since I'm writing an article in which everything else except this only relies on the property that $\mathfrak{g}$ is equipped with an $\mathrm{Ad}$-invariant scalar product, I feel like not to include the reductive assumption. But I fail to prove or find counterexamples to the above question in this case. In fact, I can't find any counterexample even for general Lie groups without any assumption.

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Addendum As noticed by Peter McNamara, we should add the assumption that $G$ is connected, since for example in the case of the real orthogonal group $O(2n)$, the adjoint action of any elements not in the identity component has eigenvalue $-1$.

Answer 1

[This is an embellished version of the answer I gave in the comment section]

The key to the solution lies in recalling that when one is playing with Lie groups, the functions that one encounters are not mere $C^\infty$ functions but are actually real-analytic. There is some discussion of this in the second edition of Knapp's book "Lie Groups Beyond an Introduction".

Now for $g\in G$, let $P_g(t)$ denote the charcteristic function of $\operatorname{Ad}(g)$. Then the function $g\mapsto P_g(-1)$ is an analtyic function on $G$. Since this function is nonzero at the identity, it is nonzero on a dense subset of the identity component of $G$. (so we have an affirmative answer for connected $G$.)

As is currenly noted in both the question and in the comments, for disconnected $G$, the example of the even orthogonal group limits what we can say away from the identity component.

I guess the case of $\pi_0(G)$ odd is still unresolved. Aber das ist eine andere Geschichte und soll ein andermal erzählt werden.

Answer 2

If $\pi_0(G)$ is odd, it's dense. For each nonzero connected component, for some $n$, $g \to g^{2^n}$ maps the connected component to itself. Take an element $g$ of the connected component. For some $m$, $g^{2^{nm}}$ has no $-1$ eigenvalues. (we can ignore the non-root of unity eigenvalues, and the roots of unity eventually become roots of unity of odd order, thus not $-1$.) So some element of the connected component has no $-1$ eigenvalues, so we can follow Peter McNamara's argument.

Note that these arguments work for any representation, not just the adjoint.