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ct.class idea – map of endomorphism operad

The respond isn’t any already for $Z = mathbb{R}$. The endomorphism operad of $mathbb{R}^2 cong mathbb{C}$ comprises the Lawvere idea of $mathbb{C}$-algebras, which can’t act on $mathbb{R}$. This quantities to proving the next barely peculiar assertion:

No topological $mathbb{C}$-algebra (acceptation, for our functions a $mathbb{C}$-algebra geared up with a topology such that addition, multiplication, and scalar multiplication by a hard and fast $lambda in mathbb{C}$ are all steady; importantly, we do not require continuity in $lambda$) can breathe homeomorphic to $mathbb{R}$.

*Proof.* Suppose $A$ is a topological ring homeomorphic to $mathbb{R}$. Then $(A, +)$ is a topological group homeomorphic to $mathbb{R}$. By Gleason-Montgomery-Zippin’s resolution to Hilbert’s fifth downside, $(A, +)$ is a $1$-dimensional merely related Lie group, therefore $(A, +) cong (mathbb{R}, +)$ as topological teams; to any extent further we’ll write $A = mathbb{R}$.

(**Edit:** This is overkill, as anticipated. There is a very elementary dispute {that a} topological group (not essentially abelian!) homeomorphic to $mathbb{R}$ should breathe isomorphic to $(mathbb{R}, +)$; behold this math.SE query.)

Write $cdot : mathbb{R} instances mathbb{R} to mathbb{R}$ for the multiplication on $A$. By rescaling we might occupy WLOG that the multiplicative identification is $1$. It follows that $q_1 cdot q_2 = q_1 q_2$ for rational $q_1, q_2$, after which by continuity it follows that $r_1 cdot r_2 = r_1 r_2$ for all $r_1, r_2$, so $A cong mathbb{R}$ as a topological ring.

(So $(mathbb{R}, +, instances)$ is the exclusive topological ring homeomorphic to $mathbb{R}$, which is a enjoyable incontrovertible fact that I do not cerebrate I’ve seen earlier than.)

In specific, $A cong mathbb{R}$ as a hoop, so admits no $mathbb{C}$-algebra construction (plane one the place scalar multiplication might breathe discontinuous) as a result of there is no such thing as a ring homomorphism $mathbb{C} to mathbb{R}$ (since $mathbb{R}$ doesn’t have an component satisfying $x^2 = -1$). $Box$

Here I’ve blithely ignored the dissimilarity between a morphism of Lawvere theories and a morphism of operads, however I do not cerebrate it issues: concretely, the endomorphism operad of $mathbb{R}^2 cong mathbb{C}$ comprises binary operations describing addition and complicated multiplication, and unary operations describing complicated scalar multiplication, and nullary (or unary relying on flavor) operations describing the additive and multiplicative identities, satisfying the $mathbb{C}$-algebra axioms, and a morphism into one other endomorphism operad $textbf{End}(X)$ implies equipping $X$ with the identical operations satisfying the identical axioms, and that is all we necessity (plane if there might or might not breathe different eerie issues occurring moving the diagonal and projection maps).

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