random cohomology for quantum nerds - 2019-10-23 (page 1 of 2)

Leaky Nun

04:17

so let's compute $H^2(C_n,\Bbb Z)$

$f:C_n \times C_n \to \Bbb Z$, $f(g^i g^j, g^k) + f(g^i, g^j) = g^i f(g^j, g^k) + f(g^i, g^j g^k)$

according to theory we can have $f(1,g^j) = f(g^i,1) = 0$ WLOG

$f(g^{i+1}, g) + f(g^i, g) = f(g, g) + f(g^i, g^2)$

$f(g,g^k) + f(g,1) = f(1, g^k) + f(g,g^k)$

maybe I should really do it via the projective resolution thing

$\cdots \to \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow \varepsilon \Bbb Z \to 0$
$\cdots \to \Bbb Z[G^5] \to \Bbb Z[G^4] \to \Bbb Z[G^3] \to \Bbb Z[G^2] \to \Bbb Z[G^1] \to \Bbb Z \to 0$

need to specify what $1$ goes to

$1 \in \Bbb Z$, $1 \in \Bbb Z[G]$, $(g,1) \in \Bbb Z[G^2]$

need $x \in \Bbb Z[G^3]$ with $dx = (1+g+\cdots+g^{n-1})(g,1) = (N,N)$

$x = (N,N,N)$, $dx = -(N,N) + (N,N) - (N,N) = -(N,N)$

so up to signs it works

I don't like this sign anyway, I like $d(g_0, g_1, \cdots, g_n) = \sum_{i=0}^n (-1)^i (\widehat{g_i})$

wait then I need $(1,g)$ instead of $(g,1)$

so now $1 \in \Bbb Z$, $1 \in \Bbb Z[G]$, $(1,g) \in \Bbb Z[G^2]$, $(N,N,N) \in \Bbb Z[G^3]$

why do I get $0$ next

this is the wrong direction anyway

aha there's something wrong with me

$N(g,1) \ne (N,N)$

it doesn't even make sense

$N(g,1) = (g,1) + (g^2,g) + (g^3,g^2) + \cdots + (1,g^{n-1})$

oops I forgot it's $(1,g)$

$d(1,g) = g-1$

$dN(g,1) = Nd(g,1) = N(g-1) = 0$

$N(1,g) = (1,g) + (g,g^2) + \cdots + (g^{n-1},1)$

$d(1,g,g^2) = (g,g^2) - (1,g^2) + (1,g)$

yeah why is this exact

ah they prove that this is contractible

this is not very constructive

ok we have $ds+sd=1$

if $dx = 0$ then $dsx = x$

wait this is very constructive

ok they constructed $s$ by $s(g_0, \cdots, g_n) := (1, g_0, \cdots, g_n)$

$d(1,1,g) = (1,g) - (1,g) +(1,1)$ wait this doesn't work

Leaky Nun

05:15

wait maybe this works

$d(1,g^i,g^{i+1}) = (g^i,g^{i+1}) - (1,g^{i+1}) + (1,g^i)$

yeah this works

let's do the opposite direction

$\cdots \to \Bbb Z[G^5] \to \Bbb Z[G^4] \to \Bbb Z[G^3] \to \Bbb Z[G^2] \to \Bbb Z[G^1] \to \Bbb Z \to 0$
$\cdots \to \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow \varepsilon \Bbb Z \to 0$

0: $1 \mapsto 1$

1: $(1,g^i) \mapsto x$ such that $(g-1)x = d(1,g^i)$, so $x = \frac{g^i-1}{g-1}$

2: $(1,g^i,g^{i+j}) \mapsto x$: $d(1,g^i,g^{i+j}) = g^i(1,g^j) - (1,g^{i+j}) + (1,g^i)$

so we need $Nx = g^i \frac{g^j-1}{g-1} - \frac{g^{i+j}-1}{g-1} + \frac{g^i-1}{g-1}$

check that $(g-1)(\cdots) = g^i(g^j-1) - (g^{i+j}-1) + (g^i-1) = 0$

ah this makes a difference when $i+j \ge n$

so for $i+j \ge n$ we should have $x=1$ up to sign, and otherwise $x=0$

and this is the 2-cocycle right

$f(g^i,g^j) = \begin{cases} 1 & i+j \ge n \\ 0 & i+j<n \end{cases}$

now $H^2(C_n,M) = M/N(M)$

10 hours ago, by MatheinBoulomenos

$\cup:C^n(G,A) \times C^k(G,B) \to C^{n+k}(G,A \otimes B)$
$(f \cup f')(g_1, \dots, g_n,g_{n+1}, \dots, g_{n+k})=f(g_1, \dots, g_n) \otimes g_1g_2 \dots g_n f'(g_{n+1}, \dots, g_{n+k})$

10 hours ago, by MatheinBoulomenos

as a special case of Tate-Nakayama, if you fix a generator of $\alpha \in H^2(C_n,\Bbb Z)$ (which is just cyclic of order $n$ again), then for all $C_n$-modules $M$ and all $k \in \Bbb Z$, the map $\beta \mapsto \beta \cup \alpha:\widehat{H}^k(C_n,M) \to \widehat{H}^{k+2}(C_n,M)$ is an isomorphism

no, $H^2(C_n,M) = M^G/NM$

so given $\beta \in M^G$ we want to define $f:C_n\times C_n\to M$

$f(g^i,g^j) = \begin{cases} \beta & i+j \ge n \\ 0 & i+j<n \end{cases}$

we need to check that $f(g^i g^j, g^k) + f(g^i, g^j) = g^i f(g^j, g^k) + f(g^i, g^j g^k)$

which should work

MatheinBoulomenos

09:28

@LeakyNun how does that contradict anything I said?

$\Bbb Z$ carries the trivial action here

MatheinBoulomenos

09:51

If you want to compute $H^2(C_n,Z)$, consider the SES $0 \to \Bbb Z \to \Bbb Z \to \Bbb Z/n\Bbb Z \to 0$ of trivial $G$-modules. This induces a LES on cohomology groups and using that $H^1(G, A) =\operatorname{Hom}(G, A) $ if the action is trivial you can easily compute $ H^2(C_n,\Bbb Z) $

Leaky Nun

14:45

@MatheinBoulomenos no I was correcting myself

9 hours ago, by Leaky Nun

now $H^2(C_n,M) = M/N(M)$

9 hours ago, by Leaky Nun

no, $H^2(C_n,M) = M^G/NM$

$G/[G,G] \to I_G/I_G^2$

$s \mapsto s-1$

@MatheinBoulomenos is there a conceptual proof that $H^1\operatorname{Hom}_G(\Bbb Z,-^G) \cong \operatorname{Hom}(G,-^G)$?

and what I'm saying is $H^1\operatorname{Hom}_G(\Bbb Z,\operatorname{Hom}_G(\Bbb Z,-)) \cong \operatorname{Hom}(G,\operatorname{Hom}_G(\Bbb Z,-))$

and I'm not sure how tensor works

at least I can't convert the right hand side into a tensor since $G$ is non-abelian

in the meantime let's revisit how Galois descent is Morita equivalence

we have two categories: $L[G,1]$-Mod and $K$-Mod

$V \mapsto V^G$, $W \mapsto L \otimes_K W$

$L$ is an ($L[G,1]$, $L$)-bimodule

now we claim that $L[G,1] \cong \operatorname{End}_K(L)$ where the forward map is tautological

maybe $V^G = L \otimes_{L[G,1]} V$

that makes no sense

by Watts theorem, if it is a tensor, it would be $L[G,1]^G \otimes_{L[G,1]} V$

I don't think even $L[G,1]^G$ makes sense but let's see

actually it might make sense

$G$ acts on $L[G,1]$ by composition: $\sigma \cdot f = \sigma \circ f$

so in terms of $\operatorname{Hom}(L,L)$ that's acting on the right factor only

this is troublesome

wait, then we can use the trivial action on $L$

so it's $\operatorname{Hom}(L_t, L)$

and we can use the famous equation $\operatorname{Hom}(L_t,L)^G = \operatorname{Hom}_G(L_t,L)$

by choosing a basis we have $\operatorname{Hom}(L_t,L)^G = \operatorname{Hom}_G(K,L)^n$

so $(L^G)^n$, so $K^n$

18 mins ago, by Leaky Nun

maybe $V^G = L \otimes_{L[G,1]} V$

so we're back to this, where $G$ acts trivially on $L$

I should be more careful with the non-canonical basis choosing

$\operatorname{Hom}(L_t,L)^G = \operatorname{Hom}_G(L_t,L) = \operatorname{Hom}(L_t,L^G) = \operatorname{Hom}(L,K)$

yeah it's $L^\ast$ instead

so $V^G = L^\ast \otimes_{L[G,1]} V$?

you know what, $V^G = \operatorname{Hom}_{L[G,1]}(L,V)$

right?

$v \mapsto (b \mapsto bv)$

$\varphi \mapsto \varphi(1)$

$\varphi((\sum a_\sigma \sigma) b) = (\sum a_\sigma \sigma) \varphi(b)$

maybe I should take the invariant quotient

how do I do that

I need to quotient by the ideal generated by $\sigma - 1$

does this even make sense

$\sum a_\sigma \sigma \mapsto \sum a_\sigma$

congratulations all you get is $L$

let $\varphi_v(b) = bv$

we need $\varphi_v(\sigma b) = \sigma \varphi_v(b)$

this seems false

$\sum a_\sigma \sigma \mapsto \sum a_\sigma$ isn't $L[G,1]$-linear

if you quotient by a left ideal you don't get a ring right

do we even have $V = \operatorname{Hom}_{L[G,1]}(L[G,1], V)$

so the rule is that if $M$ is an $(R,S)$-bimodule and $N$ is an $(R,T)$-bimodule then $\operatorname{Hom}_R(M,N)$ is an $(S,T)$-bimodule?

$s\varphi t: m \mapsto \varphi(ms)t$

$((s_1 s_2)\varphi)(m) = \varphi(m s_1 s_2) = (s_1(s_2 \varphi))(m)$

looks right

$(\sigma - 1)b = \sigma(b) \sigma - b$

$\varphi(\sum a_g g) = \sum a_g g(v) = \sum a_g v$

let's check $V = \operatorname{Hom}_{L[G,1]}(L[G,1], V)$

this might be troublesome

$\varphi_v(\sum a_g g) = \sum a_g g(v)$

ok this works

we need $(\sum a_g g)(b) = \sum a_g g(b)$

that's where I went wrong

$L[G,1]$ don't act trivially on $L$

so indeed $V^G = \operatorname{Hom}_{L[G,1]}(L,V)$

now what does this look like in terms of $\operatorname{End}_K(L)$

$L$ is a left $\operatorname{End}_K(L)$-module

$V^G = \operatorname{Hom}_{\operatorname{End}_K(L)}(L,V)$

$v \mapsto (b \mapsto bv)$

4 mins ago, by Leaky Nun

so indeed $V^G = \operatorname{Hom}_{L[G,1]}(L,V)$

I haven't checked the opposite direction

$g\varphi(1) = \varphi(g(1)) = \varphi(1)$

ok checked

$v \mapsto \varphi_v$ where $\varphi_v(b) := bv$

then $\varphi_v (\sum a_g g(b)) = \sum a_g g(b) v = \sum a_g g(b) g(v) = \sum a_g g(bv) = (\sum a_g g) \varphi_v(b)$

but that's not enough

we need to be able to project from $V$ to $V^G$

so we need a map $\operatorname{Hom}_{L[G,1]}(L[G,1],V) \to \operatorname{Hom}_{L[G,1]}(L,V)$

by Yoneda this means we need $f: L \to L[G,1]$

such that $f((\sum a_g g) b) = (\sum a_g g) f(b)$

i.e. we need $f(g(b)) = g \cdot f(b) = g(f(b)) \cdot g$ and $f(b_1 b_2) = b_1 \cdot f(b_2)$

this looks impossible

how is this done in the char 0 case where we can just take $v \mapsto \frac1n \sum gv$

take $\varphi \in \operatorname{Hom}_{L[G,1]}(L[G,1],V)$

then $\varphi' \in \operatorname{Hom}_{L[G,1]}(L,V)$ is given by $\varphi'(b) = b \frac1n \sum g\varphi(1) = b \varphi(\frac1n \sum g)$

so for $V=L[G,1]$ and $\varphi = 1$ we have $\varphi'(1) = \frac1n \sum g$

oh ok this isn't impossible

Leaky Nun

17:01

quam multa fieri non posse prius quam sunt facta iudicantur -- Naturalis Historia, Pliny the Elder

suppose $f(1) = \sum b_h h$

then we need $f(g) = g \sum b_h h = \sum g(b_h) (g \circ h)$

projection means $\varepsilon f(1) = 1$

so $\sum b_h = 1$

wait that's not the end of it

so $f(\sum a_g g) = \sum a_g g(b_h) (g \circ h)$

I don't see what goes wrong yet

oh what is $L[G,1]^G$ again

$g\sum b_h h = \sum g(b_h) (g \circ h) = \sum g(b_{g^{-1}h}) h$

so we need $b_h = g(b_{g^{-1}h})$

so $b_h = h(b_1)$

so $\sum h(b) h$ for any $b \in L$

MatheinBoulomenos

@LeakyNun there might be, but I don't know it

it's really easy to do explicitly with cochains though

Leaky Nun

that's the opposite of conceptual

MatheinBoulomenos

okay

but at least it's not tiresome

Leaky Nun

boom roasted

MatheinBoulomenos

the only proof for Hilbert 90 is by explicit computation with cocycles as well ...

Leaky Nun

17:15

I suppose

MatheinBoulomenos

you use Dirichlet's theorem on independecne of characters yeah

additive Hilbert 90 is more conceptual than multiplicative Hilbert 90

Leaky Nun

how?

MatheinBoulomenos

because of this guy:

Leaky Nun

Ben Shapiro??

MatheinBoulomenos

just kidding

we need Shapiro's lemma

Leaky Nun

17:18

ok

Arnold Samuel Shapiro (1921, Boston, Massachusetts – 1962, Newton, Massachusetts) was an American mathematician known for his eversion of the sphere. He also was the author of an article on Clifford algebras and periodicity with Raoul Bott, later redone by Michael Atiyah and Bott.

so we need $b \in L$ with $\sum_\sigma \sigma(b) = 1$

guess what

for $\Bbb C/\Bbb R$ this becomes $(a+bi)+(a-bi) = 1$ so $a = \frac12$

this doesn't look good

so for $\Bbb F_4/\Bbb F_2$ we have $0+0=0$, $1+1=0$, $a+(a+1)=1$, hey $a$ exists

@MatheinBoulomenos do you know the answer?

MatheinBoulomenos

If $f:R \to S$ is a ring homomorphism and suppose that this makes $S$ into a projective right $S$-module, then $\mathrm{Ext}^i_R(M, \mathrm{Res}_S^RN)=\mathrm{Ext}^i_S(\mathrm{Ind}_R^SM,N)$

so it's a derived version of the adjunction between extension and restriction of scalars

do you see how this implies additive Hilbert 90?

@LeakyNun answer to what?

Leaky Nun

@MatheinBoulomenos whether there is always b with sum g(b) = 1

MatheinBoulomenos

if you have a finite separable extension, then the trace is surjective

Leaky Nun

oh no

how do you prove it?

I think you use injectivity

which makes it non constructive

MatheinBoulomenos

what

Leaky Nun

17:29

the trace pairing L -> L* is injective

so it is surjective

after that you still need a basis to pick a surjective element of L*

MatheinBoulomenos

no there's a very explicit proof

Leaky Nun

oh what

MatheinBoulomenos

@LeakyNun ?

I just meant that the trace L->K is surjective

Leaky Nun

I know

MatheinBoulomenos

but you can also show that the trace pairing is not denegerate

Leaky Nun

17:31

what’s your explicit proof?

MatheinBoulomenos

you just write down a matrix for the bilinear form and compute the determinant

Leaky Nun

stares at the word matrix

MatheinBoulomenos

Let $a_1, \dots, a_n$ be a basis for $L$ over $K$. Let $\sigma_1, \dots, \sigma_n$ be the embeddings $L \to \overline{K}$.
Then we have the matrix $(\mathrm{Tr}(a_ia_j))$

thats $\sum_{k=1}^n \sigma_k(a_ia_j)=\sum_{k=1}^n \sigma_k(a_i)\sigma_k(a_j)$

which implies that if we let $A=(\sigma_i(a_j))$, then $A^TA=(\mathrm{Tr}(a_ia_j))$

Leaky Nun

d e d e k i n d

MatheinBoulomenos

you can do that

you could also note that if $a_i$ is a normal basis, then $A$ is a permutation matrix

Leaky Nun

17:40

normal basis theorem is also non constructive right

MatheinBoulomenos

no wait, that's not correct

yeah just use Dedekind

or do V A N D E R M O N DE

let $\alpha$ be a primitive element and choose the basis $1,\alpha, \alpha^2, \dots, \alpha^{n-1}$

Leaky Nun

H1(G,L) = H1(G,Hom(Z[G],K)) = H1(1,K) = 0?

@MatheinBoulomenos what’s wrong?

MatheinBoulomenos

if $a_i$ is a normal basis, then $A$ won't be a permutation matrix

Leaky Nun

oh right

MatheinBoulomenos

any primitive element + Vandermonde works

@LeakyNun uhm, I'd write $K \otimes_{\Bbb Z} \Bbb Z[G]$ but sure

I wrote $\mathrm{Ind}_R^SM$ for $S \otimes_R M$

probably bad choice

not sure what a good notation for extension of scalars is

Leaky Nun

17:47

yeah Ind is ok

MatheinBoulomenos

@LeakyNun note that this implies that $H^i(G,L)=0$ for all $i>0$

that's stronger than in the multiplicative casse

Leaky Nun

oh nice

MatheinBoulomenos

it's just Shapiro+Normal basis you don't have to actually do any computations

Leaky Nun

I’ll just note that there are no nice proofs that tr(b)=1

MatheinBoulomenos

why

I like the proof I gave

Leaky Nun

17:50

you chose a basis

MatheinBoulomenos

I even choose a primitive element

so what?

Leaky Nun

I dont like choice

MatheinBoulomenos

sometimes you have no choice but to choose

don't be so indecisive

Leaky Nun

lol

char 0 is clear, char 2 is Artin—Schreier

hmm

MatheinBoulomenos

what

how is char 2 Artin-Schreier? the statement is not just true for degree 2 extensions

Leaky Nun

17:54

if the degree is odd then tr(1)=1

MatheinBoulomenos

if it is degee 4?

Leaky Nun

then factor it into 2 degree 2

Sylow theory yay

MatheinBoulomenos

no

Leaky Nun

what

what next, S3 dont have an index 3 normal subgroup?

MatheinBoulomenos

Let $L/K$ be a Galois extension with Galois group $A_4$ (you can always construct this in any characteristic, for any finite group). Then $L/L^{A_3}$ is a degree 4 extension with no intermediate extension, because $A_3 \subset A_4$ is a maximal subgroup

and it's separable, so the trace is non-degenerate

Leaky Nun

17:57

wait what

whats Gal(L/L^A3)

MatheinBoulomenos

it's not Galois

$A_3$ is not normal in $A_4$

Leaky Nun

I thought L/E is always galois

Gal(L/L^A3) = A3

i win

MatheinBoulomenos

okay you're right I'm dumb

now look at $L^{A_3} /K $

Leaky Nun

all my extensions are galois

MatheinBoulomenos

why?

you don't need it for the statement to hold

Leaky Nun

18:00

g a l o i s descent

idk

MatheinBoulomenos

I mean, the non-denegenary of the trace pairing is interesting and useful indepedently from that

but okay

now char 3?

Leaky Nun

no clue

anyway

I guess there isn’t an explicit formula that gives 1/2 for C/R and a for F4/F2

so I’m happy with your choice

c h o i c e

so suppose $tr(b) = 1$

(yay I'm back to my computer)

then $V \to V^G: v \mapsto \sum_\sigma \sigma(b v)$ is a projection

MatheinBoulomenos

18:19

2

Leaky Nun

post this on the main chat

MatheinBoulomenos

@LeakyNun why do you need a projection $V \to V^G$? What exactly are you trying to do

Leaky Nun

@MatheinBoulomenos because that's the Morita equivalence $\operatorname{End}_K(L)-\mathsf{Mod} \implies K-\mathsf{Mod}$

you find an appropriate basis and project to the first coordinate

MatheinBoulomenos

?

just take $\mathrm{Hom}_{L(G,1)}(L,-)$

Leaky Nun

the standard Morita equivalence $\operatorname{Mat}(n,K)\mathsf{-Mod} \implies K\mathsf{-Mod}$ is $V \mapsto E_{11} V$ right

where $E_{11}$ is the matrix with only 1 at the top left hand corner

MatheinBoulomenos

18:23

is it?

Leaky Nun

do you have another equivalence?

MatheinBoulomenos

$K^n$ is a $M_n(K),K$-bimodule

just take $\mathrm{Hom}_{M_n(K)}(K^n,-)$

Leaky Nun

oh is this how you do it?

interesting

I mean surely our functors are naturally isomorphic

but your functor is more conceptual

so you're claiming that $\mathrm{Hom}_{M_n(K)}(K^n,-^n) = -$

MatheinBoulomenos

it's also way easier to see why under $\mathrm{End}_K(L) \cong L(G,1)$ we have $\mathrm{Hom}_{L(G,1)}(L,-)=(-)^G)$

Leaky Nun

why?

MatheinBoulomenos

18:28

$L \subset L(G,1)$ is a subring

so $\mathrm{Hom}_{L(G,1)}(L,V) \subset \mathrm{Hom}_{L}(L,V)$

the latter is canonically isomorphic to $V$ via $f \mapsto f(1)$

now you can easily check that $f \in \mathrm{Hom}_L(L,V)$ is contained in $\mathrm{Hom}_{L(G,1)}(L,V)$ iff $f(1) \in V^G$

Leaky Nun

yeah that works

@MatheinBoulomenos and now what's your functor $K\mathsf{-Vec} \implies M_n(K)\mathsf{-Mod}$?

MatheinBoulomenos

$K^n\otimes_K -$

Leaky Nun

so $-^n$

:P

so now why $\mathrm{Hom}_{M_n(K)}(K^n,K^n \otimes_K -) = -$?

that's like the wrong direction of the tensor-Hom adjunction

MatheinBoulomenos

there's a wrong-direction tensor-Hom adjunction if you assume that a certain one of the modules is f.g. projective

Leaky Nun

18:43

what

how

what's the adjunction?

MatheinBoulomenos

Let $A$ and $B$ be rings. let $M$ be a $(A,B)$-bimodule that is projective and finitely generated as a $A$-module. Obviously projective+finitely generated implies finitely presented. This means that $\mathrm{Hom}_A(M,-):A\text{-Mod} \to B\text{-Mod}$ commutes with finite colimits (as $M$ is projective) and filtered colimits (as $M$ is f.p.), thus it commutes with all colimits as every colimit is the filtered colimit over all subcolimits.

Now apply the Eilenberg-Watts theorem to the functor $\mathrm{Hom}_A(M,-)$

this implies that if we put $M^*=\mathrm{Hom}_A(M,A)$ (which is a $B,A$-bimodule), then $\mathrm{Hom}_A(M,-) \cong M^* \otimes -$

now in our situation we can just say that $\mathrm{Hom}_{M_n(K)}(K^n,K^n \otimes_K -) \cong \mathrm{Hom}_{M_n(K)}(K^n,K^n) \otimes_K -$

but $\mathrm{Hom}_{M_n(K)}(K^n,K^n)=K$

Leaky Nun

oh no

MatheinBoulomenos

really elementary and explicit, I know

Leaky Nun

that's exactly what I'm thinking

MatheinBoulomenos

Eilenberg-Watts is a cool theorem, though

Leaky Nun

18:55

might you have a cool equivalence $M_n(K)\mathsf{-Mod} \implies M_m(K)\mathsf{-Mod}$?

MatheinBoulomenos

take $M_{n \times m}(K)$ which is a $M_n(K),M_m(K)$-bimodule

it works just like it does for $m=1$

Leaky Nun

oh what

MatheinBoulomenos

@LeakyNun do you want to see my favorite proof that $K$ and $M_n(K)$ are Morita equivalent? It works even without mentioning the word "bimodule"

Leaky Nun

go right ahead

MatheinBoulomenos

okay, so we start by generalizing things, obviously

you know the group algebra $K[G]$, it also works if $G$ is just a monoid

Leaky Nun

18:59

sure

MatheinBoulomenos

but a monoid is just a one-object category

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds