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Stream: theory: mathematics

Topic: 2 definitions for a k-algebra agree?

Jack Jia (Jul 02 2025 at 16:35):

Let $k$ be a field. I used to assume the 2 definitions of a $k$-algebra agree: First one, a monoid $(A,\mu,\eta)$ in the symmetric monoidal category $\mathsf{Vect}_k$; second one, a ring map $\eta: k \rightarrow A$ such that $\eta(k) \subseteq Z(A)$, where $Z(A)$ is the center of $A$ (for clarity, every ring and algebra is unital and associative and I don't assume commutativity).

Now it is clear we can define multiplications in $A$: $a \cdot b = \mu(a \otimes b)$ and the unit is $\eta(1)$, so this part of equivalence has no problem, however I cannot find a way to prove $\eta(k) \subseteq Z(A)$ using the monoid definition. Moreover, I don't know if this is even true: We can regard the quaternions $\mathbb{H}$ as a $\mathbb{C}$-vector space, and I believe this will make $\mathbb{H}$ a monoid; but the centre of $\mathbb{H}$ is only $\mathbb{R}$, so the inclusion does not define an algebra if we are using definition 1.

Am I misunderstanding anything, or do these 2 definitions not agree in general?

Joe Moeller (Jul 02 2025 at 16:48):

Starting with the monoid definition you define $\eta(r):=r\cdot 1$. This is in the center because $(r\cdot 1)y=r\cdot y=y(r\cdot 1)$. Or am I missing something?

Jack Jia (Jul 02 2025 at 17:04):

@Joe Moeller

I am not sure why $r\cdot y=y (r \cdot 1)$. Also, maybe this is where I got it wrong but how is $y(r \cdot 1)$ defined? Are you using the bimodule structure on $A$? I assumed the left and right unitor in $\mathsf{Vect}_k$ are both scalar multiple on the left but maybe it was to be understood as this bimodule structure where you act on left and right?

Joe Moeller (Jul 02 2025 at 17:12):

I’m using concatenation for the monoid operation and cdot for scalar multiplication. And in general we have $y(r\cdot x)=r\cdot (yx)$. Bimodule is a fancy name for vector space in this scenario.

Mike Shulman (Jul 02 2025 at 17:14):

I wouldn't say a vector space is the same as a bimodule; a bimodule might have different left and right actions even when the ring is commutative.

Jack Jia (Jul 02 2025 at 18:16):

@Joe Moeller

Sorry, but I still don't quite understand this. Why do we have $y(r \cdot x)=r \cdot (xy)$? I don't see how that follows from the axioms, specifically what allows us to put $y$ after $x$?

Also, can you point out what's wrong with the quaternion counter example?

Jack Jia (Jul 02 2025 at 18:18):

Mike Shulman said:

I wouldn't say a vector space is the same as a bimodule; a bimodule might have different left and right actions even when the ring is commutative.

I am very confused now-- can you explain to me what the right unitor is? Is it not scalar multiplication on the left?

Mike Shulman (Jul 02 2025 at 18:39):

I think that was a typo, probably Joe meant $y(r\cdot x) = r \cdot (yx)$.

Mike Shulman (Jul 02 2025 at 18:40):

But when $x=1$ we have $y1=y=1y$.

Mike Shulman (Jul 02 2025 at 18:42):

What's wrong with the quaternions is exactly this: the multiplication is not a map $\mathbb{H}\otimes_{\mathbb{C}}\mathbb{H} \to\mathbb{H}$ because for instance $j(i\cdot 1) \neq i \cdot (j 1)$.

Mike Shulman (Jul 02 2025 at 18:42):

Jack Jia said:

can you explain to me what the right unitor is?

What do you mean by "unitor"?

Joe Moeller (Jul 02 2025 at 18:58):

Mike Shulman said:

I think that was a typo, probably Joe meant $y(r\cdot x) = r \cdot (yx)$.

Yes, sorry. Fixed.

Jack Jia (Jul 02 2025 at 19:09):

Mike Shulman said:

Jack Jia said:

can you explain to me what the right unitor is?

What do you mean by "unitor"?

By "right unitor" I meant the canonical natural iso $V \otimes I \cong V$ in a monoidal category $(\mathcal{C}, \otimes, I)$. In $\mathsf{Vect}_k$, is this still the scalar multiplication of $V$ regarded as a left $k$-vector space?

Mike Shulman (Jul 02 2025 at 19:22):

Yes. Since a field $k$ is commutative, any left module can also be regarded as a right module, and hence as a bimodule. My point was that not every bimodule is of this form.

Jack Jia (Jul 02 2025 at 20:53):

Mike Shulman said:

Yes. Since a field $k$ is commutative, any left module can also be regarded as a right module, and hence as a bimodule. My point was that not every bimodule is of this form.

So it's actually the right action, but in this case they coincide? For example, if I change the monoidal category to $\mathsf{Alg}_R$, then a monoid will need to be an $R$-bimodule, is this correct?

Mike Shulman (Jul 02 2025 at 20:57):

If $R$ is a commutative ring, then the symmetric monoidal categories of left $R$-modules and of right $R$-modules are isomorphic.

Mike Shulman (Jul 02 2025 at 20:57):

For any ring $R$, there is also a non-symmetric monoidal category of $R$-$R$-bimodules.

John Baez (Jul 02 2025 at 20:59):

Jack Jia said:

We can regard the quaternions $\mathbb{H}$ as a $\mathbb{C}$-vector space, and I believe this will make $\mathbb{H}$ a monoid;

There's no complex algebra (= monoid object in the monoidal category of complex vector spaces) whose underlying real algebra is $\mathbb{H}$.

There are different ways to prove this. The quickest way is to notice that such a complex algebra would need to be 2-dimensional (i.e. have an 2-dimensional underlying complex vector space), and every 2-dimensional algebra over a field is commutative.

To see the latter fact, notice that every 2-dimensional algebra $A$ over a field $k$ has a basis $1, x$ for some $x \in A$. Thus element in such an algebra is of the form $a1 + bx$ for $a,b \in k$, and all such elements commute.

(But perhaps you're not seeing why all such elements commute?)

Jack Jia (Jul 02 2025 at 23:14):

John Baez said:

Jack Jia said:

We can regard the quaternions $\mathbb{H}$ as a $\mathbb{C}$-vector space, and I believe this will make $\mathbb{H}$ a monoid;

There's no complex algebra (= monoid object in the monoidal category of complex vector spaces) whose underlying real algebra is $\mathbb{H}$.

There are different ways to prove this. The quickest way is to notice that such a complex algebra would need to be 2-dimensional (i.e. have an 2-dimensional underlying complex vector space), and every 2-dimensional algebra over a field is commutative.

To see the latter fact, notice that every 2-dimensional algebra $A$ over a field $k$ has a basis $1, x$ for some $x \in A$. Thus element in such an algebra is of the form $a1 + bx$ for $a,b \in k$, and all such elements commute.

(But perhaps you're not seeing why all such elements commute?)

Thanks for the explanation! I think I see that they commute since $1$ and $x$ both commute with $x$.

Jack Jia (Jul 02 2025 at 23:40):

Thanks @Joe Moeller and @Mike Shulman for the detailed explanations! I think I understand now

John Baez (Jul 03 2025 at 03:35):

Jack Jia said:

For example, if I change the monoidal category to $\mathsf{Alg}_R$, then a monoid will need to be an $R$-bimodule, is this correct?

This is a somewhat strange question since you're not saying what $R$ is, and you're acting like we know what $\mathsf{Alg}_R$ is. If $R$ is a noncommutative ring, people don't talk about $\mathsf{Alg}_R$, at least not very often.

If $R$ is a noncommutative ring the category of left $R$-modules is not monoidal in a natural way, nor is the category of right $R$-modules. But the category of $R, R$-bimodules is monoidal. A monoid in here is not usually called an $R$-algebra, because when $R$ is commutative it's not the same as the usual concept of $R$-algebra for commutative $R$. I've seen it called an $R$-ring or an $R$-bimodule algebra, and I find the latter term to be clearer.

(Note: I said if $R$ is a noncommutative ring the category of left $R$-modules is not monoidal in a natural way. But of course we can sometimes choose a monoidal structure on this category! For example there's a standard way to do this if $R$ is a bialgebra. I don't want to get into this too much, but a good example arises when $R$ is a group algebra $k[G]$ for some field $k$ and some group $G$. Then the category of left $R$-modules gets a monoidal structure because it's equivalent to the category of representations of the group $G$ on vector spaces over $k$, and this category has a well-known monoidal structure. A monoid in the category of left $R$-modules with this monoidal structure is the same as a $k$-algebra with an action of $G$ as automorphisms.)