Topological connected eccentrics, not homeomorphic to commutative Lie groups

Question

An eccentric is a universal algebra $\ (X\ \sigma\ \lambda\ \rho)\ $ such that operations $\ \sigma\ \lambda\ \rho\,:\,X\times X\to X\ $ satisfy:

1. $\quad \forall_{x\ y\,\in X}\quad \lambda(\sigma(x\ y)\ y)\ =\ x; $
2. $\quad \forall_{x\ y\,\in X}\quad \rho\,(x\ \sigma(x\ y))\ =\ y; $

Let $\ \mathcal T\ $ be a topology in $\ X\ $ such that its three operations are continuous. Furthermore, let topological space $\ \mathbf X:=(X\,\ \mathcal T)\ $ be Hausdorff and connected.

Question 1:   can $\ \mathbf X\ $ be a manifold that is not homeomorphic to any Lie group? Or can $\ \mathbf X\ $ be homeomorphic to a non-commutative Lie group?

Question 2:   can $\ \mathbf X\ $ be homeomorphic to Hilbert cube or Knaster pseudo-arc?

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Examples:

Topological spaces $\,\ \Bbb R^A\times(\Bbb R/\Bbb Z)^B,\ $ for arbitrary sets $\ A\ B\ $ (possibly empty), are manifolds which admit Abelian group structure; all such Abelian topological groups are topological eccentrics. When $\ A=\emptyset\ $ we get compact manifolds called tori.

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EDIT

see my EDIT in Weirdos but algebraic.

This thread is ok as it is but in fact, I meant also to have properties:

• $\ \forall_{a\ b\,\in\,X}\quad \sigma(\lambda(b\ a)\ a)\ =\ b;$
• $\ \forall_{a\ b\,\in\,X}\quad \sigma(a\ \rho(a\ b))\ =\ b.$

Answer 1

These are many questions...

Let me call this E-structure.

The first part of Question 1 has a positive answer. Indeed, the 7-sphere admits a magma structure $(x,y)\mapsto xy$ for which left and right multiplications are invertible with continuous inverse (octonion multiplication). Hence it admits and E-structure with $\lambda(x,y)=xy^{-1}$ and $\rho(x,y)=x^{-1}y$. It's known not to admit any topological group structure however.

[The second part of Question 1 is trivial, since every topological group is an E-structure with these laws. Maybe you mean an W-structure, i.e. the main law satisfying the medial quasigroup property $(xy)(zt)=(xz)(yt)$, as suggested by your other post.]