MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the introduction to the reprint of "Metric spaces, generalized logic and closed categories" Lawvere talks about the following situation:

Let $\mathbb R_+$ denote $\mathbb R_{\geq 0}^\infty$. Every lax monoidal endofunctor $\lambda:\mathbb R_+\to\mathbb R_+$ gives rise to an endofunctor $\lambda.:\mathrm{Met}\to\mathrm{Met}$ on the category of generalized metric spaces, again leading to the notion of $\lambda$-Lipschitz-continous maps $$ Lip^\lambda(X,Y):=Lip^1(X,\lambda.Y).$$ These are exactly the maps $f:X\to Y$ satisfying $d(x,x')\geq \lambda(d(fx,fx'))$. Lawvere goes on and "suggest a whole family of monoidal structures [on the category of metric spaces] interpolating between" the sum- (aka tensor product) and the max- (aka cartesian product) metric on the product of the underlying sets. He finally relates this to the $$\frac{1}{p}+\frac{1}{q}=1$$ business occuring in analysis.

Now for the question: Has this been worked out somewhere (replacing $\mathbb R_+$ by an arbitrary moinoidal category $\mathcal V$)?

Remark: Concerning the monoidal structures my first idea was to define various adjoints to the hom-like functors $$Lip^\lambda(X,-)$$ but as we don't have $$Lip^\lambda\times Lip^\lambda\to Lip^\lambda$$ but rather $$Lip^\lambda\times Lip^\mu\to Lip^{\mu\circ\lambda}$$ i suspect we should define $X\otimes_\lambda^\mu Y$ by an expression like $$Lip^\lambda(X,Lip^\mu(Y,Z))=:Lip^\lambda(X\otimes_\lambda^\mu Y,Z)$$ (of maybe $Lip^1$ on the right hand side). So instead of various monoidal structures we'd get various tensor products, compatible in some way...

share|cite|improve this question
What is the question? What would you like to see worked out? – Martin Brandenburg Dec 21 '11 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.