Timeline for What other monoidal structures exist on the category of sets?
Current License: CC BY-SA 3.0
8 events
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| Mar 8, 2014 at 11:18 | history | edited | John Baez | CC BY-SA 3.0 |
"singlet state" should have been "singleton".
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| Jan 30, 2014 at 19:26 | comment | added | Qiaochu Yuan | Oh, I see. $q$ doesn't even need to be mentioned in this argument; you can just start with an analytic bifunctor $\bigsqcup_{i, j} S_{i, j} \times A^i \times B^j$ and set $A, B$ to the singleton. For the singleton to be the identity I guess other ideas are required (I would be very surprised if this construction came close to exhausting monoidal structures on $\text{Set}$). | |
| Jan 30, 2014 at 9:44 | comment | added | JMAA | @QiaochuYuan Sorry, my mistake (now corrected). From the first equation you set $x=y=1$ in order to find the second equation (which holds whether you're using the sum or multiplicative group laws), which gives you that only one such $s_{i,j}$ is non-zero (and thus equal to unity and must be such $i=j=\alpha$). Then returning to general $x$ and $y$ the only remaining question is which $\alpha$ are possible. | |
| Jan 30, 2014 at 9:41 | history | edited | JMAA | CC BY-SA 3.0 |
Corrected mistake whereby RHS of second equation was incorrect
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| Jan 29, 2014 at 22:47 | comment | added | Qiaochu Yuan | Wait, I don't understand this argument at all. How can you conclude $1 = \sum s_{i, j} x^i y^j$? | |
| Jan 29, 2014 at 20:39 | history | edited | JMAA | CC BY-SA 3.0 |
Added note that changing to a multiplicative group law doesn't affect the argument or conclusion.
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| Jan 29, 2014 at 20:15 | comment | added | JMAA | @QiaochuYuan in which case the argument runs just as above and leads to the same conclusion, just that this time we find $q(x)q(y)=q((xy)^\alpha)$ rather than the same case with $f$. | |
| Jan 29, 2014 at 18:42 | history | answered | JMAA | CC BY-SA 3.0 |