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.

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.

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 $-1$.

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 $$\{ g\in G| \mathrm{ker}(1+\mathrm{Ad}_g)=\{0\} \}$$ dense in $G$?

(don't know why I can't enter braces now)

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.

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.