Two Cryptomorphic Puzzles

August 9, 2015

Posted by John Baez

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Here are two puzzles. One is from Alan Weinstein. I was able to solve it because I knew the answer to the other. These puzzles are ‘cryptomorphic’, in the vague sense of being ‘secretly the same’.

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Puzzle 1. By ‘vector space’ let’s mean a finite-dimensional real vector space. Given orientations on vector spaces $A$ and $B$ there is a standard choice of orientation on their direct sum $A \oplus B$ , which however depends on the fact that $A$ comes first and $B$ comes second. This choice makes the usual isomorphism

$(A \oplus B) \oplus C \cong A \oplus (B \oplus C)$

orientation-preserving. Can we go further, and also choose an orientation on the tensor product of any two oriented vector spaces in such a way that the usual isomorphisms

$A \otimes (B \oplus C) \cong (A \otimes B) \oplus (A \otimes C)$

$(A \oplus B) \otimes C \cong (A \otimes C) \oplus (B \otimes C)$

are orientation-preserving?

Puzzle 2. What would rings be like if we didn’t require that addition commute?

More precisely: what can you say about sets that have an group structure under addition and a monoid structure under multiplication, in such a way that multiplication distributes over addition on the right and left?

Posted at August 9, 2015 7:21 AM UTC

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Re: Two Cryptomorphic Puzzles

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In response to the second puzzle: such structures would have commutative addition regardless, as we will have that (1 + a)(1 + b) equals both 1 + a + b + ab and 1 + b + a + ab, depending on the order in which one applies left and right distributivity; cancelling the 1 and ab terms by the group structure of addition, we are left with a + b = b + a.

Posted by: Sridhar Ramesh on August 9, 2015 5:15 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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Right! Note that this proof requires that we cancel terms on both sides in

$1 + a + b + a b = 1 + b + a + a b$

So, whenever we have a set that’s a cancellative monoid under addition and a monoid under multiplication, with multiplication distributing over the addition on both the left and right, addition must be commutative.

If we drop the ‘cancellative’ condition, can we get examples where addition is not commutative? Probably.

So, it seems that for the definition of a general ‘rig’ we need to explicitly state that addition is commutative. A rig is a commutative monoid under addition and a monoid under multiplication, with multiplication distributing over the addition on both the left and right. And here I mean that multiplication distributes over $n$ -ary addition

$a \left(\sum_{i = 1}^n b_i\right) = \sum_{i=1}^n a b_i, \qquad \left(\sum_{i = 1}^n a_i\right)b = \sum_{i=1}^n a_i b$

for all $n = 0 , 1, 2, \dots$ . This is equivalent to multiplication distributing over nullary addition:

$a 0 = 0 , \qquad 0 a = 0$

and binary addition:

$a(b + c) = a b + a c, \qquad (a + b)c = a b + a c$

In the cancellative case, if multiplication distributes over binary addition it automatically distributes over nullary addition:

$a(0 + 0) = a0 + a0 \implies a0 = 0$

but otherwise I think we need to assert the nullary case separately.

Puzzle 3. Is there an interesting example of a set that’s a monoid under addition and a monoid under multiplication, with multiplication distributing over the addition on both the left and right, in which addition is not commutative?

Posted by: John Baez on August 10, 2015 2:01 AM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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Re Puzzle 3: I haven’t thought this through, but how about this? Given ordered sets $X$ and $Y$ , you can form:

their lexicographic product $X \otimes Y$ , whose elements are pairs $(x, y)$ ordered by defining $(x_1, y_1) \leq (x_2, y_2)$ whenever either $x_1 \lt y_1$ or ( $x_1 = x_2$ and $y_1 \leq y_2$ );
a kind of sum $X \oplus Y$ (generalizing the standard operation of ordinal sum); this is the disjoint union of $X$ and $Y$ , with $X$ and $Y$ retaining their usual ordering, but also declaring that $x \leq y$ whenever $x \in X$ and $y \in Y$ .

If $X$ and $Y$ are totally ordered then so are both $X \otimes Y$ and $X \oplus Y$ , so we could do all this for only totally ordered sets if we wanted to. Both $\otimes$ and $\oplus$ are assocative and unital, but neither is commutative (or cancellative). Whether $\otimes$ distributes over $\oplus$ on either side, I don’t know.

Posted by: Tom Leinster on August 10, 2015 2:09 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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If I’m not mistaken, the Puzzle-3-structures still have $x+x+y+y=(1+1)(x+y)=x+y+x+y$ , and I think this rules out Tom’s example. I asked Mace4, and it found a 3-element example, but it’s probably not an interesting one.

Posted by: Sam on August 10, 2015 5:14 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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I can’t tell if Puzzle 1 is so obvious now that people are being polite by not solving it, or if it’s so unobvious that people don’t have anything to say about it. The solution to Puzzle 2 gives the solution to Puzzle 1.

Posted by: John Baez on August 12, 2015 11:08 AM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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The concern keeping me from announcing a negative answer to Puzzle 1 due to Puzzle 2, as seemed to be what was being led to, is as to the cancellability of direct sum of oriented vector spaces. I didn’t think about it very much but it wasn’t immediately obvious to me, though it may be to others, and may become so to me.

Posted by: Sridhar Ramesh on August 12, 2015 8:19 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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Although perhaps I was overthinking it. I suppose transferring the Puzzle 2 proof, we get that the canonical iso between $1 \oplus A \oplus B \oplus (A \otimes B)$ and $1 \oplus B \oplus A \oplus (A \otimes B)$ would have to be orientation-preserving, and this reduces to the canonical iso between $A \oplus B$ and $B \oplus A$ being orientation-preserving, which doesn’t hold, thus yielding the desired negative answer.

(I reserve the right to change my mind as to the validity of what I just said in a minute)

Posted by: Sridhar Ramesh on August 12, 2015 8:28 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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Right! If both vector spaces $A$ and $B$ are odd-dimensional, the canonical isomorphism

$A \oplus B \cong B \oplus A$

is orientation-reversing, since it comes from an odd permutation of basis elements. So, the canonical isomorphism

$\begin{array}{ccl} \mathbb{R} \oplus A \oplus B \oplus (A \otimes B) &\cong & (\mathbb{R} \oplus A) \otimes \mathbb{R} \; \oplus \; (\mathbb{R} \oplus A) \otimes B \\ & \cong & (\mathbb{R} \oplus A) \otimes (\mathbb{R} \oplus B) \\ & \cong & \mathbb{R} \otimes (\mathbb{R} \oplus B) \; \oplus \; A \otimes (\mathbb{R} \oplus B) \\ & \cong & \mathbb{R} \oplus B \oplus A \oplus (A \otimes B) \end{array}$

cannot be orientation-preserving.

Posted by: John Baez on August 13, 2015 10:37 AM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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(In retrospect, my cancellability concerns were silly and misguided. But I suppose that’s often the way: confusion in the moment, clarity later.)

Posted by: Sridhar Ramesh on August 13, 2015 6:20 PM | Permalink | Reply to this

Re: Two Cryptomorphic Puzzles

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You were just wanting the analogy between Puzzle 1 and Puzzle 2 to be a bit stronger than it actually is. In Puzzle 2 we don’t need to use cancellability to get our hands on the isomorphism

$A \oplus B \cong B \oplus A$

In this situation, unlike in Puzzle 1, we already know that this isomorphism exists.

But we could make the analogy stronger. Suppose we have a category $C$ that is monoidal in two ways, called $\otimes$ and $\oplus$ . Suppose $\otimes$ distributes over $\oplus$ on both sides. And suppose the monoidal category $(C, \oplus,0)$ is actually ‘groupal’: that is, monoidal, but where every object $A$ has a dual object $-A$ where the unit and counit

$i_A: 0 \to A \oplus -A$

$e_A : -A \oplus A \to 0$

are actually isomorphisms!

Then we can use the argument you’ve given to get isomorphisms

$b_{A,B}: A \oplus B \; \cong \; B \oplus A$

Puzzle 4. Do these isomorphism make $(C, \oplus, 0)$ into a braided monoidal category?

I don’t know the answer.

Posted by: John Baez on August 17, 2015 9:45 AM | Permalink | Reply to this

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August 9, 2015