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I am an optical engineer, so please forgive any ignorance my questions betoken. I am interested in whether one can tear down the manifold of a finite dimensional Lie group,

Example 1: Take the additive group $\left(\mathbb{R},\cdot +\right)$, it is its own Lie algebra. As the lie algebra I denote it $\mathbf{g}$ and the Lie group $G$ - arising from the two topologies are isomorphic, their topologies are homoemorphic.

Example 2: I believe, if the following is right, that connected centreless Lie group must also "behave itself" in the above way, i.e. if you can give it a topology to make it into a connected Lie group in many different ways, the Lie groups so built must all have the same Lie algebra, fundamental group and connected components. For suppose two different topologies / manifolds make it into connected Lie groups $G$ and $G^\prime$; here we also allow for there being different Lie algebras $\mathbf{g}$ and $\mathbf{g}^\prime$. We think of smooth paths
$\sigma\left(\tau\right)=\exp\left(\tau X\right) \in G;\; X \in \mathbf{g}$ and $\sigma^\prime\left(\tau\right)=\exp^\prime\left(\tau X^\prime\right) \in G^\prime; \; X \in \mathbf{g}^\prime$. As for Example 1, there are two different exponential maps $\exp$ and $\exp^\prime$. In the centreless case, the adjoint $G$ and $G^\prime$ are locally isomorphic to (have the same Lie algebras as) the groups of big Ad matrices acting through the adjoint group representations on their respective Lie algebras. Now, even though $\sigma^\prime$ might be totally disconnected in $G$'s toplogy and contrawise, $\sigma, \sigma^\prime$ induce smooth paths $\exp_M\left({\rm ad}_X\right) \in \left(\mathbf{g}\right)$, $\exp_M\left({\rm ad}_X^\prime\right)\in gl\left(\mathbf{g}^\prime\right)$ through their respective adjoint representations and here $\exp_M$ is now the same, in the sense its being the restriction from the continuous one defined on $gl\left(N, K\right)$), continuous matrix exponential. Here $N$ is the Lie group dimension (if we allow for it to be different for the two topologies, we assume it is the bigger number) and $K = \mathbb{R}, \mathbb{C}$, as appropriate. So now we have potentially different linear Lie subgroups $G_L$ and $G^\prime_L$ of $gl\left(N, K\right)$, each locally isomorphic to $G$ and $G^\prime$, respectively. So we now find the universal covers of the four groups $G$, with its locally isomorphic $G_L$ and $G^\prime_L$ with its locally isomorphic $G^\prime_L$, we have linear $\hat{G} \cong \hat{G_L}$, $\hat{G}^\prime \cong \hat{G^\prime_L}$; the covers are central extensions with the discrete fundamental groups as kernel of the projections back tooriginal groups. With $G$ and $G^\prime$ being the same as sets, if the universal covers in the two topologies are not the same, one must be a cover of the other. But this would imply that they both have isomorphic Lie algebras. So the topologies can't be different (otherwise one couldn't be a universal cover). Hence the two universal coversare isomorphic as Lie groups. Modding out the centres, i.e. the fundamental group to get back to centreless groups, we show that an automorphism of G that preseres Liealgebras and fundamental groups takes it to $G^\prime$.

Example 3: I don't believe example 2 is affected by the presence of a discrete centre. Thus groups with discrete centre only can only be given a Lie group structure in essentially one way.

Example 4: Special cases of example 3 would be any connected group with semisimple Lie group.

Example 5: I believe I can get a vector generalisation of the technique in Example 1 to work for connected Abelian Lie groups.

The above leaves out animals like the Heisenberg group, other nilpotent beasts, solvable groups as could-be problem children.

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Can a Lie group as an abstract group be given more than one topology making it a Lie group?

I am an optical engineer, so please forgive any ignorance my questions betoken. I am interested in whether one can tear down the manifold of a Lie group, leaving an abstract group, and then give the group another manifold structure that makes it again into a Lie group and get anything essentially different in the process. I know the manifold replacement can be done in some special cases. Here are the examples I have been thinking about and could come up with answers to. In all examples I can make progress on, the Lie groups arising from the different manifolds built on the same set are isomorphic (as Lie groups that is: i.e. they have the same Lie algebra, fundamental group and connected components) - they are of course isomorphic as abstract groups!!!. I have a hunch that this is generally true, otherwise one might get weird examples where an abstract group might have several different Lie algebras, and I'm sure I'd have read about that somewhere by now! If someone knows anything about the general case, I'd be most interested to hear about it.

Example 1: Take the additive group $\left(\mathbb{R},\cdot +\right)$, it is its own Lie algebra. As the lie algebra I denote it $\mathbf{g}$ and the Lie group $G$ - the exponential map is the identity map. Now take one of the everywhere discontinuous, bijective solutions $\phi$ of the Cauchy functional equation $f\left(x\right) + f\left(y\right) = f\left(x + y\right)$ as detailed in the end of Chapter 2 of Hewitt and Stromberg "Real and Abstract Analysis"; $\phi$ is now an exponential map from the $g$ onto a new Lie group $G^\prime$, to wit the same group $\left(\mathbb{R},\cdot +\right)$ but now given a new topology generated by images $\phi\left(U\right)$ of open intervals in $\mathbb{R}$. This topology is of course totally disconnected in the group topology for $G$; however $\phi$ is continuous, indeed $C^\infty$ when thought of as a map from $g$ to $G^\prime$, the latter with the new topology. Indeed, the Lie groups arising from the two topologies are isomorphic, their topologies are homoemorphic.

Example 2: I believe, if the following is right, that connected centreless Lie group must also "behave itself" in the above way, i.e. if you can give it a topology to make it into a connected Lie group in many different ways, the Lie groups so built must all have the same Lie algebra, fundamental group and connected components. For suppose two different topologies / manifolds make it into connected Lie groups $G$ and $G^\prime$; here we also allow for there being different Lie algebras $\mathbf{g}$ and $\mathbf{g}^\prime$. We think of smooth paths
$\sigma\left(\tau\right)=\exp\left(\tau X\right) \in G;\; X \in \mathbf{g}$ and $\sigma^\prime\left(\tau\right)=\exp^\prime\left(\tau X^\prime\right) \in G^\prime; \; X \in \mathbf{g}^\prime$. As for Example 1, there are two different exponential maps $\exp$ and $\exp^\prime$. In the centreless case, the adjoint $G$ and $G^\prime$ are locally isomorphic to (have the same Lie algebras as) the groups of big Ad matrices acting through the adjoint group representations on their respective Lie algebras. Now, even though $\sigma^\prime$ might be totally disconnected in $G$'s toplogy and contrawise, $\sigma, \sigma^\prime$ induce smooth paths $\exp_M\left({\rm ad}_X\right) \in \left(\mathbf{g}\right)$, $\exp_M\left({\rm ad}_X^\prime\right)\in gl\left(\mathbf{g}^\prime\right)$ through their respective adjoint representations and here $\exp_M$ is now the same, in the sense its being the restriction from the continuous one defined on $gl\left(N, K\right)$), continuous matrix exponential. Here $N$ is the Lie group dimension (if we allow for it to be different for the two topologies, we assume it is the bigger number) and $K = \mathbb{R}, \mathbb{C}$, as appropriate. So now we have potentially different linear Lie subgroups $G_L$ and $G^\prime_L$ of $gl\left(N, K\right)$, each locally isomorphic to $G$ and $G^\prime$, respectively. So we now find the universal covers of the four groups $G$, with its locally isomorphic $G_L$ and $G^\prime_L$ with its locally isomorphic $G^\prime_L$, we have linear $\hat{G} \cong \hat{G_L}$, $\hat{G}^\prime \cong \hat{G^\prime_L}$; the covers are central extensions with the discrete fundamental groups as kernel of the projections back to original groups. With $G$ and $G^\prime$ being the same as sets, if the universal covers in the two topologies are not the same, one must be a cover of the other. But this would imply that they both have isomorphic Lie algebras. So the topologies can't be different (otherwise one couldn't be a universal cover). Hence the two universal covers are isomorphic as Lie groups. Modding out the centres, i.e. the fundamental group to get back to centreless groups, we show that an automorphism of G that preseres Lie algebras and fundamental groups takes it to $G^\prime$.

Example 3: I don't believe example 2 is affected by the presence of a discrete centre. Thus groups with discrete centre only can only be given a Lie group structure in essentially one way.

Example 4: Special cases of example 3 would be any connected group with semisimple Lie group.

Example 5: I believe I can get a vector generalisation of the technique in Example 1 to work for connected Abelian Lie groups.

The above leaves out animals like the Heisenberg group, other nilpotent beasts, solvable groups as could-be problem children.

I have a University of Pittsburgh 2007 PhD. Thesis "On the Uniqueness of Polish Group Topologies" by Bojana Pejic that I believe may be relevant, but so far have not made great headway in understanding it. If someone could point me to other material on this question, I'd be grateful.