23,063 reputation
360167
bio website wwwmath.uni-muenster.de/u/…
location Münster, Germany
age 26
visits member for 4 years, 3 months
seen 11 hours ago

PhD student interested in the interactions between algebraic geometry and category theory. More specifically, I "model" algebraic geometry on cocomplete symmetric monoidal categories.



Email: [my last name] [at] uni-muenster.de


11h
awarded  Enlightened
13h
awarded  Nice Answer
21h
revised Universal property of module categories over monads
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1d
revised Universal property of module categories over monads
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1d
comment Universal property of module categories over monads
Exactly, Tom. My motivation for the terminology "modules" is the example $T=A \otimes -$ for some algebra $A$ in a monoidal category, here $T$-modules are left $A$-modules (not $A$-algebras). In my understanding an algebra should carry some sort of associative binary operation, but a $T$-action is a morphism $Tx \to x$ satisfying two conditions, which really abstracts the definition of a left module.
1d
asked Universal property of module categories over monads
1d
comment Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$
The condition in the def. of $P$ should be $p = 0 \Rightarrow N \geq 1$, right?
1d
comment Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$
@Piotr: $H^{-1}$ cannot be written like that. I think the normal form is rather $\mathbb{N} \sqcup \{H^p + n : p \in \mathbb{Z} \setminus \{0\}, n \in \mathbb{N}\}$.
2d
revised Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$
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2d
asked Semiring of vector bundles on $\mathbb{C}\mathbb{P}^1$
Apr
10
awarded  Popular Question
Apr
4
awarded  Taxonomist
Apr
4
revised Categorification of the integers
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Apr
4
asked Categorification of the integers
Mar
31
comment Analogy between the exterior power and the power set
Meanwhile I've been able to formalize the analogy, using "$\pm$-enriched" categories.
Mar
30
accepted Rank vanishing in tensor categories
Mar
30
comment Rank vanishing in tensor categories
I am interested in char. $0$. If $\mathcal{L}$ is an invertible object, then we don't necessarily have $\mathrm{rk}(\mathcal{L})=1$? In fact, the rank equals the signature of $\mathcal{L}$, which is an involution of $1$, and $-1$ is possible e.g. when considering $\mathbb{Z}$-graded objects of $\mathcal{C}$ (twisted symmetry). If $X$ is $1_\mathcal{C}$ concentrated in degree $1$, then every dualizable graded object has the form $M=\sum_n M_n \otimes X^{\otimes n}$ with $M_n$ dualizable (almost all $0$), $\mathrm{rk}(X)=-1$, hence $\mathrm{rk}(M) = \sum_n (-1)^n \mathrm{rk}(M_n)$. Correct?
Mar
30
comment Rank vanishing in tensor categories
Sorry, forgot to mention that everything is over $\mathbb{Q}$ (as in Deligne, section 7). Now it's included.
Mar
30
revised Rank vanishing in tensor categories
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Mar
29
revised Rank vanishing in tensor categories
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