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group dynamics – Broccoli games: or, how do you deal with unfun sessions before you get to have fun again?

This may be a silly question, but I’m still quite curious. How do deal with unfun game sessions that you feel are still kind of required?

The context of my question is, our group just had, effectively, a TPK. We made a few tactical errors but more pressing were horrendous dice rolls. Like, our Blessed One critically botched her parry attempt, stabbed herself and went from full HP to dying… in a low-damage game system. We’re kind of used to bad luck on our part and NPCs rolling really well, so it wasn’t a shocker. It still leaves us with either a game over, or a capture scenario. And thinking about playing through one of those almost makes me prefer the TPK.

The campaign has, up to that point, been rife with moments where we PCs felt powerless and incompetent. We were stuck in a tournament where each event had several NPCs that far outleveled us, or plain cheated. We won exactly one discipline, and only because we cheated after the GM told us how. Our greatest success came from an NPC on our side heroically defeating the NPC antagonist (the prince of the realm who is set up to become a major enemy later on) in the joust. We tried to prove an innocent man not guilty and got everything turned around on us until we did exactly what our GM told us to do. Long story short, we sucked, but we’re starting characters so we accepted that we sucked. Sucking is natural at this stage. Plus I feel we players are a little stupid.

After the tournament, we were following an escaped murderer through the wilderness, got into a fight we could’ve easily won but lost instead. And now I’m worried we’re right back to being helpless and getting humiliated by the NPCs. And sure, this could all be a setup for a daring escape… but at this point, I don’t trust us to do that right, and even if we do, that means having to sit through the prison scenes first. We lost, so I guess some punishment is in order before we can try being heroes again? Eating our broccoli even if we hate it.

So, what’s your approach when you know or suspect the next few sessions will be unfun, but need to happen so you can get back to having fun? Do you have any tips? Should I talk to the GM about giving us a summary of what happened? Our GM is one of my best, dearest friends, I don’t want to upset him.

SharePoint user group permission list

How to generated an excel spreadsheet for User group permission for a specific site?

algebraic topology – Why is there no (tame) knot group has a presentation with deficiency two

I am recently reading Rolfsen’s Knots and Links, 3E6 says there is no (tame) knot group which has a presentation with deficiency two, and I am wondering why.

What I know:

  1. Every tame knot group (in $mathbb{R}^3$ or $S^3$) has a finite presentation of deficiency one, by Wirtinger presentation.
  2. The abelinization of each knot group is infinite cyclic, by either Wirtinger presentation for tame knots or abelinizing it to get $H_1$ and apply Mayor-Vietorus.

My question is pretty simple:
Intuitively with each one of the relation in a group presentation, we can get rid of one generator at most, so abelinizing a finite group presentation can reduce the number of generators by at most the number of the relations. Since this gives $mathbb{Z}$, we have the deficiency of any presentation of the knot group at most one. How can I make this rigorous though?

Thanks in advance for helpful ideas.

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cohomology – Higher Chow group without condition of admissibility for cycles

Here is my motivation. I am interested in calculation of the Higher Chow group of a field. By Totaro result Chow group $CH^n(F,n)$ is isomorphic to Milnor $K$-theory. I am trying to generalise this result to the group $CH^n(F,n)$. That is to prove Goncharov conjecture in this degree. The solution of A. Goncharov’s conjecture about so-called emph{strong reciprocity law} allows to present the group $CH^{n-1}(F,n)$(G means Goncharov) as cohomology of some complex which is very similar to cubical Higher chow group.(Complex $N_l$ below) But problem is that in such a presentation there is no condition of admissibility for cycles. But for Higher Chow group there is.

So roughly speaking, is it possible to present the group $CH^{n-1}(F,n)otimes mathbb Q$ without condition of admissibility for cycles? Below I give precise conjecture. The idea is that the differential in the cubical higher chow group is very similar to tame-symbol map. But the tame symbol is defined in any case. Below I give exact statement.

Let $k$ be an algebraically closed field of characteristic zero. Denote by $mathcal c F_d$ the set of isomorphism classes of fields of transcendence degree $d$ over $k$.

For a field $F$ and non-negative integers $k,l$ denote by $N_{k,l}'(F)$ the quotient of the vector space $left(Lambda^{k+l} F^timesright)otimes mathbb Q$ by the following elements $c_1wedge dots wedge c_lwedge f_{l+1}wedge dots wedge f_{l+k}, f_iin F, c_iin k$. Denote by $N_{k,l}(F)$ the coinvariants $N_{k,l}(F)=(N_{k,l}(F))_{Aut(F)}$.

Consider the following complex called $N_l$ placed in degrees $(l-2,l)$:
$$bigopluslimits_{Fin mathcal F_2}N_{2,l}(F)xrightarrow{d_{N_l,l-2}} bigopluslimits_{Fin mathcal F_1}N_{1,l}(F)xrightarrow{d_{N_l,l-1}} (Lambda^l k^times)otimes mathbb Q.$$

The differential in this complex is obtained by the sum of the tame-symbol map over all the divisorial valuations of the field $F$. So the differental $d_{N_l,l-1}$ is the sum of tame symbols over all points of the corresponding curve. Do you know anything about construction like this?

Denote by $z^l(k,n)$ the cubical higher Chow complex given by codimension $l$ admissible cycles in $(mathbb P^1backslash{1})^{2l-n}$. There is a natural morphism of complexes $mathcal T_lcolon tau_{geq n-1}(z^l(k, *))to tau_{geq n-1}(N_l)$. This morphism maps a cycle $C^lin (mathbb P^1backslash{1})^{2l-n}$ to the element $xi_1wedge dots wedge xi_{2l-n}$, where $xi_i$ is the restriction of the $i$-th projection $(mathbb P^1backslash{1})^{2k-n}tomathbb P^1$ to $C^l$. One can show that this morphism is well-defined.

It follows from Suslin reciprocity law and Totaro theorem that this morphism is isomorphism in degree $l$. Here is my main question:

is it isomorphism in degree $l-1$?

As I said above another motivation is that $H^{l-1}(N_l)$ is isomorphic to Goncharov’s cohomology group.

Here is the exact definition of the differential $d_{N_l}$.

For $Fin mathcal F_d$ denote by $dval(F)$ the set of discrete(with values in $mathbb Z$) divisorial valuations. (A valuation is divisorial if it is given by a Cartier divisor on some birational model of $F$). For $Fin mathcal F_d$ and $nuin dval(F)$ there is the tame-symbol map:
$$partial_nucolon N_{k,l}'(F)to N_{k-1,l}'(overline {F_nu}).$$

The map $d_{N_l,l-1}$ is defined by the formula:

$$d_{N_l,l-1}(F)(x)=sumlimits_{nuin dval(F)}partial_nu(x).$$

Using resolution of singularity arguments one can show that for all but finite number of $nu$ we have $d_{N_l,l-1}(F)(x)=0$.

To construct map $d_{N_l,l-2}$ we need for any $F_2in mathcal F_2, F_1in mathcal F_2$ define the map $N_{2,l}(F_2)to N_{1,l}(F_1)$. It is defined by the formula

$$xmapstosumlimits_{substack{nuin dval(F_2)\overline{F_{2,nu}}cong F_1}}partial_nu(x).$$

The fact that $N_l$ is a complex is a similar to so-called Parshin reciprocity law.

Breadcrumb showing username instead of group name in drupal 8.x

I have created a group and reached to members adding page. The displaying breadcrumb is below.

I have install group 8.x 1.3 module
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Proposed resolution

username should be replaced with the group name.
Home > [Group name] >> Content >> Add >> Group membership

ag.algebraic geometry – Picard group of moduli of principal bundles

I am looking for the Picard group of moduli space of principal $G$-bundles for a connected reductive complex algebraic group $G$. Is it isomorphic to $mathbb{Z}$?
If not, what can we say when $G=mathrm{Sp}(2n,mathbb{C})?$

Is there any reference for this?

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