# Tag Info

7

All your examples are perfectly fine. Another example is the set of $(n\times n)$-matrices equipped with multiplication. This multiplication is not commutative, but we say two matrices commute iff $AB=BA$. Or we could say "Matrix multiplication is commutative on diagonal matrices".

6

It is not incorrect to refer to a non-commutative operation as being commutative when operating on certain elements. In fact, this notion is very important! For instance, the center of a group is the set of elements that commute with every element of the group. So the answer is a resounding "no". The algebraic structure need not be generally commutative to ...

3

I think that everything you've said is fine. It is trivially true that for any operation, $x$ and $y$ commute whenever $x=y$, so we wouldn't normally bother to say that. But the property of commuting is a local property. For example, in a group, the center of the group is the set of elements that commute with every element of the group. Since we always ...

3

Apparently it's called a (commutative) inverse monoid. For further details, see Wikipedia1, 2, 3 or Lawson's Inverse Semigroups4. (I haven't proven that the sets of axioms are equivalent. You may want to reserve the bounty for someone who does so.)

2

There is a construction of a "universal morphism" in Brown's Topology and Groupoids, chapter 8.1. We assume that $G$ is a groupoid, $\sigma:Ob(G)\to X$ is a set map. Then we can construct a groupoid $U$ whose object set is exactly $X$, and a morphism $\barσ:G\to U$ whose object function is $σ$. The idea is similar to the construction of the free product of ...

2

A theorem is an important statement in mathematics on its own, while a lemma's main/sole purpose is proving a theorem. The Pythagorean Theorem is an important result in mathematics, and we should not call it a lemma just because it has many profound consequences. Quite the contrary, in fact, that is a reason we should call it a theorem.

1

Short answer: Because it mainly deals with function spaces. As opposed to calculus, where all the norms of Euclidean spaces are equivalent (and all Hausdorff topologies which respect their linear nature), in the case of infinite dimensional linear spaces norms are not equivalent. And Functional analysis is the "analysis" study of infinite (in general) ...

1

"rationalizing the denominator" still applies. To make explicit that the process is using $k$ as its base field, "$k$-rationalizing the denominator" (or rationalizing "over $k$" / "relative to $k$"). The suggestion is therefore $\mathbb{R}$-rationalizing the denominator "Realize" and "decomplexify" have accepted meanings very different from this. ...

1

As you can see in this link (page 5, col.2), Hoerl (presumably the inventor of ridge regression) "gave the name "ridge regression" to his procedure because of the similarity of its mathematics to methods he used earlier i.e., "ridge analysis," for graphically depicting the characteristics of second order response surface equations in many predictor ...

1

Edit Based on OP Comments Actually, I realised that the sum of two bernoulli rvs with different ps will result in *under*dispersion, so perhaps underdispersed? If you don't want to be associated with dispersion models, then why not "Heterogeneous Binomial Sum", its clearer than generalized binomial, as there are several ways you could generalize it.

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