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Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.

EDIT 2: So far, in the case of a connected, compact, semi-simple Lie group, I have found out nothing interesting. It is known that these groups are symmetric spaces, or, more precisely, give rise to symmetric spaces. Namely, consider an (open) subgroup $H_0$ of the stabilzer $H(\sigma)$ of $G$, where $\sigma$ is an involutive automorphism of $G$. Then, it is known that the quotient space $G/H_0$ is a symmetric space when one uses the definition of a symmetric space via Lie theory. The question I am now asking myself is how the curvature tensor behaves in general. There is a long formula for the curvature tensor of $G$ in "Differential Geometric Structures" by Walter A. Poor. Then there is a remark about the Levi-Civita connection which takes a particular simple form when the metric is bi-invaraint, namely $\nabla_X Y \cdot_{G} Z = \frac{1}{2}[X,Y]\cdot_{G}Z$. Then the author writes: The converse is true if $G$ is connected and the curvatue tensor takes the form $R(W,X)Y=-\frac{1}{4}[[W,X],Y]$. I am not quite sure what the author wants to say here. Does he mean that then we may choose a bi-invaraint metric and do so in order to obtain this more elegant formula? (This is what I believe actually and if so, this seems like a dead end for a (possible) proof of the above statement - Any comments clarifying this linguistic problem are welcome.). I will further be occupied with the problem of Claudio.

Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.

EDIT 2: So far, in the case of a connected, compact, semi-simple Lie group, I have found out nothing interesting. It is known that these groups are symmetric spaces, or, more precisely, give rise to symmetric spaces. Namely, consider an (open) subgroup $H_0$ of the stabilzer $H(\sigma)$ of $G$, where $\sigma$ is an involutive automorphism of $G$. Then, it is known that the quotient space $G/H_0$ is a symmetric space when one uses the definition of a symmetric space via Lie theory. The question I am now asking myself is how the curvature tensor behaves in general. There is a long formula for the curvature tensor of $G$ in "Differential Geometric Structures" by Walter A. Poor. Then there is a remark about the Levi-Civita connection which takes a particular simple form when the metric is bi-invaraint, namely $\nabla_X Y \cdot_{G} Z = \frac{1}{2}[X,Y]\cdot_{G}Z$. Then the author writes: The converse is true if $G$ is connected and the curvatue tensor takes the form $R(W,X)Y=-\frac{1}{4}[[W,X],Y]$. I am not quite sure what the author wants to say here. Does he mean that then we may choose a bi-invaraint metric and do so in order to obtain this more elegant formula? (This is what I believe actually and if so, this seems like a dead end for a (possible) proof of the above statement - Any comments clarifying this linguistic problem are welcome.). I will further be occupied with the problem of Claudio.

EDIT 3: In fact, what I've posted before seems like a dead end. Now I want to check if there's something interesting in Morse-theory.

Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.

Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.

EDIT 2: So far, in the case of a connected, compact, semi-simple Lie group, I have found out nothing interesting. It is known that these groups are symmetric spaces, or, more precisely, give rise to symmetric spaces. Namely, consider an (open) subgroup $H_0$ of the stabilzer $H(\sigma)$ of $G$, where $\sigma$ is an involutive automorphism of $G$. Then, it is known that the quotient space $G/H_0$ is a symmetric space when one uses the definition of a symmetric space via Lie theory. The question I am now asking myself is how the curvature tensor behaves in general. There is a long formula for the curvature tensor of $G$ in "Differential Geometric Structures" by Walter A. Poor. Then there is a remark about the Levi-Civita connection which takes a particular simple form when the metric is bi-invaraint, namely $\nabla_X Y \cdot_{G} Z = \frac{1}{2}[X,Y]\cdot_{G}Z$. Then the author writes: The converse is true if $G$ is connected and the curvatue tensor takes the form $R(W,X)Y=-\frac{1}{4}[[W,X],Y]$. I am not quite sure what the author wants to say here. Does he mean that then we may choose a bi-invaraint metric and do so in order to obtain this more elegant formula? (This is what I believe actually and if so, this seems like a dead end for a (possible) proof of the above statement - Any comments clarifying this linguistic problem are welcome.). I will further be occupied with the problem of Claudio.

Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

Since we are dealing with symmetric spaces, in particular with compact semi-simple Lie groups, we already know that the space can be equipped with a bi-invariant metric. As you mentioned before, we know that compact semi-simple Lie groups with a bi-invariant metric form a symmetric space. However, the results on which you base your assumption were derived by heavily depending on the symmetric space structure. This seems to be a total loop for me (in the methodological sense!).

If you are referring to the general question below, I would say that I do not see the point in proving something like that. This is not meant as an offence, but I actually do not understand where this question is supposed to be going. All considerations above were made for symmetric spaces, that require the compact Lie Group to be already equipped with a bi-invaraint metric.

I know this is not quite an answer, but I don't know how to comment, so if someone of the moderators feels this should be converted in a comment, I'd not necessarily oppose to doing that.

EDIT: I am currently working on the problem. Thanks to Claudio, it has become clearer to me what actually should be shown.