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Let $G$ be a non-abelian group and $\mathcal S$ and $\mathcal T$ be group topologies on $G$. What is the largest group topology $\tau$ on $G$ with $\tau \subseteq \mathcal T\cap \mathcal S$?

In abelian case it is easy to find a base of neighborhoods around $1$ for $\tau$. In this paper there are some propositions abut infimum of two field topologies. But I could not find a general investgation about infimum of two group topologies.

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  • $\begingroup$ Can you list a case where $\mathcal T\cap \mathcal S$ is not a group topology? It seems like it might be... $\endgroup$
    – Thomas
    Commented Jul 13, 2014 at 19:17
  • $\begingroup$ see P. Samuel, Ultrafilters and compactifications of uniform spaces, Trans. Amer. Math. Soc. 64 (1948), 100-134. There's another question about this in mathoverflow, I did not find the link. But that's easy to get the point if you try to prove the intersection is a group topology. $\endgroup$
    – H. Khas
    Commented Jul 13, 2014 at 19:25

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Start with $\tau_0 = \mathcal{T} \cap \mathcal{S}$. For each successor ordinal $\alpha+1$, let $\tau_{\alpha+1}$ be the set of elements in $\tau_\alpha$ whose preimage under multiplication is in $(\tau_\alpha)^2$ (product topology) and whose preimage under inverse is in $\tau_\alpha$. For each limit ordinal $\alpha$, let $\tau_\alpha$ be the intersection of the previous topologies. This process eventually terminates; $\tau$ is the intersection of all the $\tau_\alpha$.

I don't know how many steps are required, though, or if there is a non-recursive way to find $\tau$.

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  • $\begingroup$ that may eventually become a proof for its existence. But there is simpler way to prove its existence. My question is how to specify a base of neighborhoods of $1$ for it. In abelian case a subbase is obtained by products of neghborhoods of $\mathcal T$ and $\mathcal S$. $\endgroup$
    – H. Khas
    Commented Jul 14, 2014 at 0:01
  • $\begingroup$ Won't the analog of Colin Reid's construction work equally well if, instead of working with the whole topologies $\tau_\alpha$, we work only with the neighborhood filters at the identity? If so, then it seems to answer the question. $\endgroup$
    – Andreas Blass
    Commented Jul 14, 2014 at 4:15
  • $\begingroup$ @H.Khas "that may eventually become a proof". This IS a proof - it is enough to continue the induction along an ordinal whose cofinality is larger than the number of topologies on $G$. $\endgroup$
    – Adam Przeździecki
    Commented Jul 14, 2014 at 9:32
  • $\begingroup$ @Colin Reid - I think one step is enough if you replace your demand that the inverse image is open in $(\tau_\alpha)^2$ with the demand that all inverse images in $(\tau_0)^n$ for $n\geq 2$ are open. $\endgroup$
    – Adam Przeździecki
    Commented Jul 14, 2014 at 9:37

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