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I'm trying to learn about natural transformations, but I'm lost trying to work through the details of one.

Problem: Let $Comm$ be the category whose objects are commutative rings and whose maps are ring homomorphisms, and let $M$ be the category whose objects are monoids and maps are monoid homomorphisms. Then fix $n \in \mathbb{N}$ and let $F: Comm \to M$ be the functor defined by $F(R)=M_{n}(R)$ and $G: Comm \to M$ be the functor $G(R)=R^{\star}$ (the ring $R$ under multiplication). Interpret the determinant map as a natural transformation.

Attempted proof. Since determinant is a function from a matrix to the elements of it corresponding ring, the proposed natural transformation is a family of maps in $M$ where for any matrix monoid $M_{n}(R)$ there is the determinant map to the multiplicative monoid $R^{\star}$. Since $\det(I)=1$ and $\det(AB)=\det(A)\det(B)$, we know that $\det$ is a homomorphism so there actually is a determinant map for any matrix monoid in $M$. To show this is a natural transformation, we need to shows that it commutes, so we fix a map $f: A \to B$ in $Comm$ and we need to show $\det \circ F(f)=G(f) \circ \det$. We know $F(f)$ is a map from $M_{n}(A)$ to $M_{n}(B)$ and thus $\det \circ F(f)$ is a map from $M_{n}(A)$ to $B^{\star}$. Similarly, $G(f)$ is a map from $A^{\star}$ to $B^{\star}$ and det maps $M_{n}(A)$ to $A^{\star}$ so $G(f) \circ \det$ is also a map from $M_{n}(A) \to B^{\star}$.

This is the point where I get a little lost. It seems to me that the way to show these two maps are equal is to fix some $X\in M_{n}(A)$ and show that both homomorphisms send $X$ to the same element of $B^{\star}$. For the right hand side, we have $G(f) \circ \det (A)=G(f) (\det( A))=f(\det(A))$. This looks confusing to me, but if I understand correctly, the functor $G$ sends a map/homomorphism $f$ in $Comm$ to the identical map/homomorphism in $M$ and the real "heart" of $G$ is how it forgets the additive structure of objects in $Comm$. Do I understand this correctly? Then, for the other map, we want to look at $\det \circ F(f) (A)$, but I don't really understand what $F(f)(A)$ is. We know $F(f)$ is some homomorphism from $M_{n}(A)$ to $M_{n}(B)$, but I don't know what homomorphism it actually is, so I'm stuck.

mle
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goatman2743
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3 Answers3

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First of all, your discussion of the functor $G$ looks good.

The objects in the category $Comm$ are commutative rings, so a morphism $f: A \to B$ is a unital ring homomorphism. The functor $F$ maps a ring to the monoid of matrices over that ring. How does this functor act on morphisms? Entry-wise! Given $X = \left( x_{ij} \right) \in M_n(A) = F(A)$, $$ F(f): M_n(A) \to M_n(B) $$ maps $$ \left( x_{ij} \right) \mapsto \left( f(x_{ij}) \right). $$

By the way, in my opinion, the multiplicative monoid is not necessarily the "heart" of a ring, as there are really two forgetful functors from $Comm$. The first is your $F$ that just keeps around the multiplicative monoid. The other just keeps around the abelian group.

Similar exercise: Is it true that with the forgetful functor $F': Comm \to Ab$, the trace map is a natural transformation?

Sammy Black
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  • Thanks! That all makes sense now. For the functor $F'$, I would say the answer is no, because in general $tr(I_{R})$ does not equal $1_{R}$. Therefore, the trace function is not a homomorphism, so there is not a trace map between $M_{n}(R)$ and $R^{+}$ in the category of monoids. – goatman2743 Mar 17 '14 at 19:49
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The point is that determinant is essentially given by the same algebraic formula for all rings, and ring homomorphisms respect it.

For instance, consider $n = 2$, so that $F(R) = M_2(R)$. Fix some ring homomorphism $f: A \to B$. Now let's show that $\det_B \circ F(f) = G(f) \circ \det_A$, where $\det_R: M_2(R) \to R^\star$ is obviously $\det([[a, b], [c, d]]) = ad - bc$.

Then, to show $\det_B \circ F(f) = G(f) \circ \det_A$ we need to show that for any $M \in M_2(R)$ it holds that $\det_B(F(f)(M)) = G(f)(\det_A(M))$.

Now, let $M = [[a, b], [c, d]]$. Then $F(f)(M) = [[f(a), f(b)], [f(c), f(d)]]$, and so $\det_B(F(f)(M)) = f(a)f(d) - f(b)f(c) = f(ad-bc)$, and the last equality follows from the fact that $f$ is ring homomorphism.

On the other hand, $G(f)(\det_A(M)) = G(f)(ad - bc) = f(ad-bc)$, since $G(f)$ is the same map as $f$, set-theoretically. Thus, we obtain the desired equality.

The case for $n>2$ is similar, as you'll clearly see.

xyzzyz
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Let me make a precision to what have been said.

As said before, $F(f)$ can be taken to be the application of $f$ to each of the components of the respective matrix and $G(f)$ is essentially the same $f$ but formally working on a monoid instead of a ring.

If you want to proof rigorously that the induced mappings $F(f)$ and $G(f)$ conmute with the determinant then just take the general expression of the determinant in a commutative ring, Wikipedia calls it Leibnitz's formula: $$\det(A) = \sum_{\sigma \in S_n} (sgn(\sigma) \prod_{i = 1}^n a_{i,\sigma_i})$$

You will admit that since ring homomorphism $f$ preserves sums and products, the conclusion is straight forward.

Reference

Tom Leinster, Basic Category Theory, page 30

user1868607
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