## at.algebraic topology – extended double-two-cycle conditions: mathematical structure behind it?

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1. The usual group 2 cocycle condition:

Let us remember the usual so-called homogeneous group 2 cocycle $$mu (a, b, c)$$ the cohomology group $$H ^ 2 (G, U (1))$$ Where $$U (1) = mathbb {R} / mathbb {Z}$$ is given by

$$frac { mu (b, c, d) mu (a, b, d)} { mu (a, c, d) mu (a, b, c)} = 1.$$
where everyone $$a, b, c, d in G$$,

• We can focus on the case $$G$$ is a finite group (or even a finite Abelian group if you want to simplify further.)

See references to group cohomology:

The homogeneous group 2 cocycle $$mu (a, b, c)$$ can be covered by a homogeneous group 2 cocycle
$$omega (A, B): = omega (a d ^ {- 1}, b d ^ {- 1}) = mu (a d ^ {- 1}, b d ^ {- 1}, 1).$$
So if we define $$a d ^ {- 1} = A$$ and $$b d ^ {- 1} = B$$. $$c d ^ {- 1} = C$$, then

$$frac { mu (bd ^ {- 1}, cd ^ {- 1}, 1) mu (ad ^ {- 1}, bd ^ {- 1}, 1)} { mu (ad ^ {- 1}, cd ^ {- 1}, 1) mu (ac ^ {- 1}, bc ^ {- 1}, 1)} = frac { mu (B, C, 1) mu (A, B, 1)} { mu (A, C, 1) mu (AC ^ {- 1}, BC ^ {- 1}, 1)} = frac { omega (B, C) omega (A, B)} { omega (A, C) omega (A C ^ {- 1}, BC ^ {- 1})} = 1.$$
or equivalent is the 2-Cocycle condition:
$$frac { mu (A, B, 1)} { mu (AC ^ {- 1}, BC ^ {- 1}, 1)} = frac { mu (A, C, 1)} { mu (B, C, 1)} Leftrightarrow frac { omega (A, B)} { omega (AC ^ {- 1}, BC ^ {- 1})} = frac { omega (A, C)} { omega (B, C)}.$$

1. Extended double-two-cycle condition: mathematical structure behind it?

Let's define a new object call $$F$$ which is related to the usual homogeneous group 2 cocycles $$mu_1$$ and $$mu_2$$ (also inhomogeneous group 2 cocycles $$omega_1$$ and $$omega_2$$ ) with two tensor product inputs:
$$F (A, B, alpha, beta): = mu_1 (A otimes alpha, B otimes beta, 1) = omega_1 (A otimes alpha, B otimes beta)$$
Likewise
$$F (A, B,?,?) = Μ 2 (A × B, α × β, 1) = & ohgr;$$

The 2-Cocycle condition for a homogeneous Group 2-Cocycle $$mu_1$$ (also an inhomogeneous group 2 cocycle $$omega_1$$ )
becomes:

$$frac { omega_1 (A otimes alpha, B otimes beta)} { omega_1 (AC ^ {- 1} otimes alpha gamma ^ {- 1}, BC ^ {- 1} otimes beta gamma ^ {- 1})} = frac { omega_1 (A otimes alpha, C otimes gamma)} { omega_1 (B otimes beta, C otimes gamma)}$$

$$Rightarrow boxed { frac {F (A, B, alpha, beta)} { F (AC ^ {- 1}, BC ^ {- 1}, alpha gamma ^ {- 1}, beta gamma ^ {- 1}) } = frac {F (A, C, alpha, gamma)} { F (B, C, beta, gamma) }} day 1}$$

The 2-Cocycle condition for a homogeneous Group 2-Cocycle $$mu_2$$ (also an inhomogeneous group 2 cocycle $$omega_2$$ )
becomes:

$$frac { omega_2 (A otimes B, alpha otimes beta)} { omega_2 (AC ^ {- 1} otimes B gamma ^ {- 1}, alpha C ^ {- 1} otimes beta gamma ^ {- 1})} = frac { omega_2 (A otimes B, C otimes gamma)} { omega_2 ( alpha otimes beta, C otimes gamma)}$$

$$Rightarrow boxed { frac {F (A, B, alpha, beta)} { F (AC ^ {- 1}, B gamma ^ {- 1}, alpha C ^ {- 1}, beta gamma ^ {- 1}) } = frac {F (A, B, C, gamma)} { F ( alpha, beta, C, gamma) }} day 2}$$

Here everyone $$A, B, C, alpha, beta, gamma in G$$,

My puzzle for you: Are there any known mathematical structures behind these two extended double-two-cycle conditions in Equation (1) and Equation (2)? If so, what does the corresponding co-cycle class solution look like? (Are there certain modified terms in group cohomology?

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## MySQL: I have to return rows under certain conditions

I have 3 tables: work, recording, publication

1 work can contain several recordings and 1 recording is only shown in 1 release

Table: work

``````+---------+-----------+
| work_id | name      |
+---------+-----------+
| 1       | Hello     |
| 3       | Luna      |
| 4       | Feel good |
| 5       | My self   |
+---------+-----------+
``````

TABLE: Recording

``````+---------------------------------------------------------------------+
| recording_id | work_id | release_id | name        | is_art | is_vid |
+---------------------------------------------------------------------+
| 45           | 1       | 45         | Hello4      | 1      | 0      |
| 78           | 3       | 67         | Luna5       | 1      | 0      |
| 23           | 5       | 128        | My self (r) | 1      | 0      |
| 95           | 5       | 156        | My self II  | 1      | 0      |
+---------------------------------------------------------------------+
``````

TABLE: Approval

``````+--------------------------------------------+
| release_id | name    | year | month | day  |
+--------------------------------------------+
| 45         | Yo      | 1998 | 12    | NULL |
| 67         | Testing | 1967 | 3     | 3    |
| 128        | Maybe   | 2018 | 10    | 21   |
| 156        | Again   | 2018 | 10    | NULL |
+--------------------------------------------+
``````

Basically for everyone `work`I want to give them back `recording` Where `is_art = 1` and `is_vid = 0` AND they were `release` is the oldest (oldest year, month and date). Sometimes it is `release` can have the same `year`. `month` and `day`In this case, I will probably have to find a unique identifier `release_id`

The result set should look like this:

``````+---------+---------------------------------------+
| work_id | name      | recording_id | name       |
+---------+---------------------------------------+
| 1       | Hello     | 45           | Hello4     |
| 3       | Luna      | 78           | Luna5      |
| 5       | My self   | 23           | My self (r)|
+---------+---------------------------------------+
``````

So far I've created this query, but frankly, as a newbie, I know everything is wrong. Duplicate rows are returned. I fell as I need to use it `group by` and subqueries, but after 2 days of searching and testing I can't create a solution …

``````SELECT
w.work_id, w.`name`, r.recording_id, r.`name`
FROM
work w
JOIN recording r on w.work_id = r.work_id
JOIN `release`rl ON r.release_id = rl.release_id
WHERE
r.is_art = 1 and r.is_vid = 0
ORDER BY w.work_id, rl.released_date_year, rl.released_date_month, rl.released_date_day
``````

## analytical geometry – Find the diameter of the circle for the following given conditions.

To let $$PQ$$ and $$RS$$ Tangents at the ends of the diameter $$PR$$ of a circle with radius $$r$$, If $$PS$$ and $$RQ$$ cross at one point $$X$$ you will then find the circumference of the circle $$2r$$ in terms of $$PQ, RS$$

My attempt is as follows: At a right angle $$triangle PXR$$

$$PX ^ 2 + RX ^ 2 = PR ^ 2$$
$$4r ^ 2 = PX ^ 2 + RX ^ 2 tag {1}$$

$$triangle PXQ sim triangle SXR$$therefore

$$dfrac {SX} {PX} = dfrac {RX} {QX} = dfrac {SR} {PQ}$$

$$dfrac {PS} {PX} = dfrac {SR + PQ} {PQ}$$
$$PX = dfrac {PS cdot PQ} {SR + PQ} tag {2}$$

In the same way

$$RX = dfrac {RQ cdot RS} {SR + PQ} tag {3}$$

Put all of this in equation $$(1)$$

$$4r ^ 2 (PQ + RS) ^ 2 = PS ^ 2 cdot PQ ^ 2 + RQ ^ 2 cdot RS ^ 2 tag {4}$$

Now we have to get rid of it $$PS$$ and $$RQ$$

$$PS ^ 2 = PR ^ 2 + RS ^ 2 tag {5}$$
$$RQ ^ 2 = PQ ^ 2 + PR ^ 2 tag {6}$$

Put all of this in equation $$(4)$$

$$4r ^ 2 (PQ + RS) ^ 2 = (PR ^ 2 + RS ^ 2) PQ ^ 2 + (PR ^ 2 + PQ ^ 2) RS ^ 2$$
$$4r ^ 2 (PQ ^ 2 + RS ^ 2 + 2PQ cdot RS-PQ ^ 2-RS ^ 2) = 2PQ ^ 2RS ^ 2$$
$$2r = sqrt {PQ cdot RS}$$

Now you can see that the solution took a long time. It took me some time to consider whether I miss a property here that can be useful.

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## postgresql – Postgres 11+: Are covering indexes (INCLUDE) useful for join / where conditions?

I want to better understand when index covering can be useful to enable index-only scans in Postgres 11+. As the documentation says, given the cover index

``````CREATE INDEX tab_x_y ON tab(x) INCLUDE (y);
``````

Queries like this can only use it for index scans:

``````SELECT y FROM tab WHERE x = 'key';
``````

Now I'm wondering if such a cover index could also allow index scans if the cover columns are shown as conditions. For example, assume a cover index:

``````CREATE INDEX tab_x_y_z ON tab(x) INCLUDE (y, z);
``````

Would this only allow index scans for the following queries?

``````SELECT z FROM tab WHERE x = 'key' AND y = 1;

SELECT x, y, z FROM (VALUES ('key1'),('key2'),('key3')) sub(id)
JOIN tab ON tab.x = sub.id WHERE y = 1;
``````

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## Simplification of expression under symmetry conditions

To simplify the following expression:

``````f(i1**k1)f(i2**j2)f(j1**k1p)-f(i1**j2)f(i2**k1)f(j1**k1p)
-f(i1**k1)f(i2**j1)f(j2**k1p)+f(i1**j1)f(i2**k1)f(j2**k1p)
-f(i2**j2)f(k1**i1)f(k1p**j1)+f(i1**j2)f(k1**i2)f(k1p**j1)
+f(i2**j1)f(k1**i1)f(k1p**j2)-f(i1**j1)f(k1**i2)f(k1p**j2)
+f(j2**k1p)f(k1**i1**i2**j1)+f(k1p**j2)f(k1**i1**i2**j1)
-f(j1**k1p)f(k1**i1**i2**j2)-f(k1p**j1)f(k1**i1**i2**j2)
+f(i2**k1)f(k1p**i1**j2**j1)+f(k1**i2)f(k1p**i1**j2**j1)
-f(i1**k1)f(k1p**i2**j2**j1)-f(k1**i1)f(k1p**i2**j2**j1)
``````

If the replacement of i1 by i2 or j1 by j2 changes the sign of the expression, we have more than 2 for the number of expressions in the product (… ** k1 ** k2 ** …):

``````f(...**k1**k2**..)==(-1)f(...**k2**k1**...)
``````

However:

``````f(k1**k2)==KroneckerDelta(k1,k2)-f(k2**k1)
``````

The output must be:

``````4*f(i2**j2)(f(j1**k1p)f(i1**k1)-f(k1**i1)f(k1p**j1))
-2*f(j1**j2**i2**k1p)*KroneckerDelta(i1**k1)
+2*f(k1**i1**i2**j2)*KroneckerDelta(j1**k1p)
``````

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## ac.commutative algebra – conditions on a ring that make the core of certain matrices trivial via the power series ring

To let $$R$$ be a (unital) commutative ring and leave $$varphi$$ Bean $$n times n$$ Matrix about $$R ((x))$$, Are there any conditions $$R$$ that guarantee that $$ker ( varphi) cap R (x) ^ n = 0$$ implied $$ker ( varphi) = 0$$? Of course, some conditions are required. Following the example at the very end of this document, there are $$1 times 1$$ Matrices for which the implication fails. According to sentence 5 of the same article, however, this problem is resolved if $$R$$ is assumed to be noetheric (in fact the sentence gives the much stronger implication) $$ker ( varphi) cap R = 0 Rightarrow ker ( varphi) = 0$$). This is not an answer in the event that $$varphi$$ is not $$1 times 1$$ even though.

I am not sure whether the implication is also true in this case $$R$$ is a field. I am particularly interested in the case $$R = k ((x_1, ldots, x_m))$$ for a field, $$k$$but there is no hope on this front if it doesn't work if $$R$$ is a field. If the implication is not valid under any non-degenerate conditions, the answer changes if we accept the entries from $$varphi$$ are all polynomials?

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## Solving equations – Conditions for positives and negatives – Beyond "Reduce" – How can this be attacked?

Unfortunately, the following attempt to find the conditions for the positive doesn't work. There is a solution to 0, but the negative condition is also troublesome.

``````Assuming((n | vi | Vq | Vd | Q | Bd | Bq | Subscript(b, q)) (Element) Reals && n >= 2 && vi > 0 && Vq > 0  && Vd >= 0 && Q >= 1, FullSimplify@Reduce(-((n^2 (-1 + 2 n) (Q ((Bd + Bq)^2 - vi) + vi + Vq - !(*UnderoverscriptBox(((Sum)), (j = 1), (Q))*SuperscriptBox(((Bd + bi + Bq - *SubscriptBox((b), (j)))), (2)))))/((Bd + bi + Bq)^2 (-1 + n)^2 Q + Q (Vd + vi + Vq) - 2 n Q (Vd + vi + Vq) + n^2 (vi + Vq - !(*UnderoverscriptBox(((Sum)), (q = 1), (Q))*SuperscriptBox(((Bd + bi + Bq - *SubscriptBox((b), (q)))), (2)))))^2) > 0))
``````

Any suggestions on how to tackle this problem to still find conditions? Or am I resigned to not finding general expressions?

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## Logic theorem conditions – mathematical batch exchange

We were asked to find the logical sentence:
All small triangles are left over from all squares.

∀x: small (x) ∧ triangle (x) ⇒ ∀y: square (y) ∧ left of (x, y)

However, the software simulated a case in which my judgment is not true. Apparently, if there is a small triangle but no squares, does my formula return true even though it should be untrue? Posted on Categories Articles

## dnd 5e – What are the conditions if you act twice blind / deaf to the same creature?

Your quote of the relevant rules is spot on. The effects are not stacked.

## Latest replaces previous

Relevant Sage Advice interview with Jeremy Crawford (~ 25: 30) about a person under the spell (true polymorph) and this spell is replaced by a subsequent spell of the same spell.

"The next one on the stack replaces the previous one on the stack, unless the previous one … is more powerful."

Rubiksmoose pointed out that the recent errata for the Player's Handbook clarifies:

Combine magical effects (p. 205). In the first paragraph, the following sentence was added to the first paragraph: "Or the newest effect occurs if the cast parts are of the same strength and their duration overlaps."

## The most powerful"

As there is no more specific term for the game, it fits the native meaning of potent:

With great power, influence or impact.

A stronger force or effect would be stronger. A higher spell, a higher spell DC, or a spell-enhanced effect would increase the strength of one spell over another.

## Simultaneous effects

In the event that two effects occur simultaneously on a target's turn, the player or DM responsible for the target determines the order of the effects.

For example, two wizards held blindness / deafness, and both actuated the trigger "when the fighter is moving forward" and cast their spell on the fighter. In this case, the fighter is executed simultaneously with the fighter's move. The person at the table who controls the fighter chooses which spell affects them first.

## How can I iterate through a table based on multiple conditions in SQL?

student table

``````student_id  student_name
1           John
2           Mary
``````

touch table

``````student_id   year   grade_level  school        Course    Mark
1          2015    10          Smith High     Algebra   95
1          2015    10          Smith High     English   96
1          2016    11          Smith High     Geometry  85
1          2016    11          Smith High     Science   88
2          2015    10          Smith High     Algebra   98
2          2015    10          Smith High     English   93
2          2016    11          Smith High     Geometry  97
2          2016    11          Smith High     Science   86
``````

I try to show results for each year and which class a student has given the grade.

So the final output I'm looking for looks like this:

``````[student_id1] [year1] [grade1] [school1]
[course1] [mark1]
[course2] [mark2]
[course3] [mark3]...
[course1] [mark1]
[course2] [mark2]
[course3] [mark3]...
[course1] [mark1]
[course2] [mark2]
[course3] [mark3]...
``````

This would all go in one column / row. In this particular example, this would be my result:

``````1 2015 10 Smith High
Algebra 95
English 96
1 2016 11 Smith High
Geometry 85
Science 88
2 2015 10 Smith High
Algebra 98
English 93
2 2016 11 Smith High
Geometry 97
Science 86
``````

Every time a student's ID, year, class or name changes, I have a line for it and go through the classes that are in that group. And all of that would be in one column / row.

This is what I've had so far, but I'm not sure how to properly go through the course and grades for each group. I would appreciate if I can be steered in the right direction.

``````select s.student_id + '' + year + '' + grade_level + '' + school
from students