aarjaneiro/planetmath huggingface dataset:
symmetric group on three letters

symmetric group on three letters

This example is of the

symmetric group

on

3

letters, usually denoted by

S

3

. Here, we are considering the set of

bijective functions

on the set

A

=

{

1

,

2

,

3

}

which naturally arise as the set of

permutations

on

A

. Our

binary operation

is function composition which results in a new bijective function. This example develops the table for

S

3

. We start by listing the elements of our group. These

elements

are listed according to the second method as described in the entry on

permutation notation

.

e

=

(

1 2 3

1 2 3

)

r

=

(

1 2 3

2 1 3

)

a

=

(

1 2 3

2 3 1

)

s

=

(

1 2 3

3 2 1

)

b

=

(

1 2 3

3 1 2

)

t

=

(

1 2 3

1 3 2

)

Here, our group is just

S

3

=

{

e

,

a

,

b

,

r

,

s

,

t

}

. Now we can start to multiply and then fill in the table.
First, we calculate the square of each element.

a

2

=

(

1 2 3

2 3 1

)

⁢

(

1 2 3

2 3 1

)

=

(

1 2 3

3 1 2

)

=

b

b

2

=

(

1 2 3

3 1 2

)

⁢

(

1 2 3

3 1 2

)

=

(

1 2 3

2 3 1

)

=

a

r

2

=

(

1 2 3

2 1 3

)

⁢

(

1 2 3

2 1 3

)

=

(

1 2 3

1 2 3

)

=

e

s

2

=

(

1 2 3

3 2 1

)

⁢

(

1 2 3

3 2 1

)

=

(

1 2 3

1 2 3

)

=

e

t

2

=

(

1 2 3

1 3 2

)

⁢

(

1 2 3

1 3 2

)

=

(

1 2 3

1 2 3

)

=

e

Next, we will fill in the upper right

3

⁢

x

⁡

3

block, we only need

a

⁢

b

and

b

⁢

a

since we can use the fact that there can be no repetition in any row or column.

a

⁢

b

=

(

1 2 3

2 3 1

)

⁢

(

1 2 3

3 1 2

)

=

(

1 2 3

1 2 3

)

=

e

b

⁢

a

=

(

1 2 3

3 1 2

)

⁢

(

1 2 3

2 3 1

)

=

(

1 2 3

1 2 3

)

=

e

The other

3

⁢

x

⁡

3

blocks are also

similar

. Now continuing with the upper left 3 x 3 block, we go through the table again using the fact that there can be no repetition in any row or column.

a

⁢

r

=

(

1 2 3

2 3 1

)

⁢

(

1 2 3

2 1 3

)

=

(

1 2 3

3 2 1

)

=

s

a

⁢

s

=

(

1 2 3

2 3 1

)

⁢

(

1 2 3

3 2 1

)

=

(

1 2 3

1 3 2

)

=

t

Similarly, we

complete

the final blocks of the table.

r

⁢

a

=

(

1 2 3

2 1 3

)

⁢

(

1 2 3

2 3 1

)

=

(

1 2 3

1 3 2

)

=

t

r

⁢

b

=

(

1 2 3

2 1 3

)

⁢

(

1 2 3

3 1 2

)

=

(

1 2 3

3 2 1

)

=

s

s

⁢

a

=

(

1 2 3

3 2 1

)

⁢

(

1 2 3

2 3 1

)

=

(

1 2 3

2 1 3

)

=

r

s

⁢

r

=

(

1 2 3

3 2 1

)

⁢

(

1 2 3

2 1 3

)

=

(

1 2 3

2 3 1

)

=

a

Finally, we fill in the table using the calculated values above.

e

a

b

r

s

t

e

e

a

b

r

s

t

a

a

b

e

s

t

r

b

b

e

a

t

r

s

r

r

t

s

e

b

a

s

s

r

t

a

e

b

t

t

s

r

b

a

e

Title

symmetric group on three letters

Canonical name

SymmetricGroupOnThreeLetters

Date of creation

2013-03-22 15:52:24

Last modified on

2013-03-22 15:52:24

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

13

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 20B30

Generated on Fri Feb 9 19:21:05 2018 by

LaTeXML
Cayley’s theorem

Cayley’s theorem

Let

G

be a group, then

G

is

isomorphic

to a

subgroup

of the

permutation group

S

G

If

G

is finite and of order

n

, then

G

is isomorphic to a subgroup of the permutation group

S

n

Furthermore, suppose

H

is a

proper subgroup

of

G

. Let

X

=

{

H

⁢

g

|

g

∈

G

}

be the set of

right cosets

in

G

. The map

θ

:

G

→

S

X

given by

θ

⁢

(

x

)

⁢

(

H

⁢

g

)

=

H

⁢

g

⁢

x

is a

homomorphism

. The kernel is the largest

normal subgroup

of

H

. We note that

|

S

X

|

=

[

G

:

H

]

!

. Consequently if

|

G

|

doesn’t divide

[

G

:

H

]

!

then

θ

is not an

isomorphism

so

H

contains a non-trivial normal subgroup, namely the kernel of

θ

.

Title

Cayley’s

theorem

Canonical name

CayleysTheorem

Date of creation

2013-03-22 12:23:13

Last modified on

2013-03-22 12:23:13

Owner

vitriol (148)

Last modified by

vitriol (148)

Numerical id

7

Author

vitriol (148)

Entry type

Theorem

Classification

msc 20B35

Related topic

CayleysTheoremForSemigroups

Generated on Fri Feb 9 19:21:12 2018 by

LaTeXML
empirical proof that solving opposing faces of a Rubik’s cube does not necessarily solve the middle layer

empirical proof that solving opposing faces of a Rubik’s cube does not necessarily solve the middle layer

“Is it true that if two opposite faces of a Rubik’s cube are completed the middle layer is completed too?”

Not necessarily, even though this is usually the case. A cube with two opposing faces completed but the middle layer scrambled does not arise from normal play, but can be contrived easily enough.

Given a Rubik’s cube and a solve-by-layers

algorithm

, start with a scrambled cube.

1. Choose a color to be the top face. Let’s say you choose white and the opposing face is blue.

2. Solve the top layer (i.e., making sure the colors of all the pieces on the top layer are in the right place).

3. Inspect the middle layer for any instance of the color you want on the bottom face (blue, in this example). Use Edge Mover 1 and Edge Mover 2 (moves that take a piece from the bottom layer and place it on the middle layer) to remove any blue pieces from the middle layer, but leave any other pieces in the middle layer so that each face’s middle layer has two or three different colors.

4. Use Two Corner Swap (changes position of bottom corners), Three Corner Swap (changes

orientation

of bottom corners), Edge Flipper Left and Edge Flipper Right (change position of edge pieces in the bottom layer) to solve the bottom face, or if desired, the bottom layer. Each of these moves, upon completion, leaves the top and middle layer as it was before beginning the move, but makes the appropriate change to the bottom layer.

Now you should have a cube with white on the top face, blue on the bottom face and some“wrong” colors on the middle layer.

Hopefully this empirical proof can lead to a more mathematically rigorous proof.

References

1

D. Singmaster, ”Notes on Rubik’s Magic Cube” (1980)

2

Winning Moves, Inc.,

Solutions Hints! Booklet

, (1996)

Title

empirical proof that solving opposing faces of a Rubik’s cube does not necessarily solve the middle layer

Canonical name

EmpiricalProofThatSolvingOpposingFacesOfARubiksCubeDoesNotNecessarilySolveTheMiddleLayer

Date of creation

2013-03-22 15:36:56

Last modified on

2013-03-22 15:36:56

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

9

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 20B99

Classification

msc 00A08

Generated on Fri Feb 9 19:21:21 2018 by

LaTeXML
Frobenius group

Frobenius group

A

permutation group

G

on a set

X

is

Frobenius

if no

non-trivial element

of

G

fixes more than
one element of

X

. Generally, one also makes the

restriction

that at least one non-trivial element
fix a point. In this case the

Frobenius group

is called

non-regular

.

The

stabilizer

of any point in

X

is called a

Frobenius complement

, and has the remarkable
property that it is distinct from any

conjugate

by an element not in the

subgroup

. Conversely,
if any

finite group

G

has such a subgroup, then the

action on cosets

of that subgroup makes

G

into a Frobenius group.

Title

Frobenius group

Canonical name

FrobeniusGroup

Date of creation

2013-03-22 13:16:30

Last modified on

2013-03-22 13:16:30

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

5

Author

bwebste (988)

Entry type

Definition

Classification

msc 20B99

Defines

Frobenius complement

Generated on Fri Feb 9 19:21:23 2018 by

LaTeXML
proof of Cayley’s theorem

proof of Cayley’s theorem

Let

G

be a group, and let

S

G

be the

permutation group

of the underlying set

G

. For each

g

∈

G

, define

ρ

g

:

G

→

G

by

ρ

g

⁢

(

h

)

=

g

⁢

h

. Then

ρ

g

is

invertible

with

inverse

ρ

g

-

1

, and so is a

permutation

of the set

G

.

Define

Φ

:

G

→

S

G

by

Φ

⁢

(

g

)

=

ρ

g

. Then

Φ

is a

homomorphism

, since

(

Φ

⁢

(

g

⁢

h

)

)

⁢

(

x

)

=

ρ

g

⁢

h

⁢

(

x

)

=

g

⁢

h

⁢

x

=

ρ

g

⁢

(

h

⁢

x

)

=

(

ρ

g

∘

ρ

h

)

⁢

(

x

)

=

(

(

Φ

⁢

(

g

)

)

⁢

(

Φ

⁢

(

h

)

)

)

⁢

(

x

)

And

Φ

is

injective

, since if

Φ

⁢

(

g

)

=

Φ

⁢

(

h

)

then

ρ

g

=

ρ

h

, so

g

⁢

x

=

h

⁢

x

for all

x

∈

X

, and so

g

=

h

as required.

So

Φ

is an

embedding

of

G

into its own permutation group. If

G

is finite of order

n

, then simply numbering the elements of

G

gives an embedding from

G

to

S

n

.

Title

proof of Cayley’s

theorem

Canonical name

ProofOfCayleysTheorem

Date of creation

2013-03-22 12:30:50

Last modified on

2013-03-22 12:30:50

Owner

Evandar (27)

Last modified by

Evandar (27)

Numerical id

5

Author

Evandar (27)

Entry type

Proof

Classification

msc 20B99

Generated on Fri Feb 9 19:21:27 2018 by

LaTeXML
(p,q) shuffle

(

p

,

q

)

shuffle

Definition

.

Let

p

and

q

be positive natural numbers. Further,
let

S

⁢

(

k

)

be the set of

permutations

of the numbers

{

1

,

…

,

k

}

.
A permutation

τ

∈

S

⁢

(

p

+

q

)

is a

(

p

,

q

)

shuffle

if

τ

⁢

(

1

)

<

⋯

<

τ

⁢

(

p

)

,

τ

⁢

(

p

+

1

)

<

⋯

<

τ

⁢

(

p

+

q

)

.

The set of all

(

p

,

q

)

shuffles is denoted by

S

⁢

(

p

,

q

)

.

It is clear that

S

⁢

(

p

,

q

)

⊂

S

⁢

(

p

+

q

)

. Since a

(

p

,

q

)

shuffle is
completely determined by how the

p

first

elements

are mapped, the

cardinality

of

S

⁢

(

p

,

q

)

is

(

p

+

q

p

)

.
The wedge product of a

p

-form and a

q

-form can be defined as a sum
over

(

p

,

q

)

shuffles.

Title

(

p

,

q

)

shuffle

Canonical

name

pqShuffle

Date of creation

2013-03-22 13:33:59

Last modified on

2013-03-22 13:33:59

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 20B99

Classification

msc 05A05

Synonym

shuffle

Related topic

ShuffleOfLanguages

Generated on Fri Feb 9 19:21:30 2018 by

LaTeXML
(p,q) unshuffle

(

p

,

q

)

unshuffle

Let

p

and

q

be positive natural numbers. Further, let

S

⁢

(

k

)

be the

symmetric group

on the numbers

{

1

,

…

,

k

}

. A

permutation

τ

∈

S

⁢

(

p

+

q

)

is a

(

p

,

q

)

unshuffle

if there exist

i

1

<

⋯

<

i

p

and

j

1

<

⋯

<

j

q

s.t.

τ

⁢

(

i

1

)

=

1

,

…

,

τ

⁢

(

i

p

)

=

p

and

τ

⁢

(

j

1

)

=

p

+

1

⁢

…

,

τ

⁢

(

j

q

)

=

p

+

q

.

Alternatively a

(

p

,

q

)

unshuffle

is a permutation

τ

∈

S

⁢

(

p

+

q

)

s.t.

τ

-

1

is a

(

p

,

q

)

shuffle

.

Since a

(

p

,

q

)

unshuffle is completely determined by

{

i

1

,

…

,

i

p

}

, the cardinality of

{

σ

∈

S

⁢

(

p

+

q

)

|

σ

is an unshuffle

}

is

(

p

+

q

q

)

.

Title

(

p

,

q

)

unshuffle

Canonical

name

pqUnshuffle

Date of creation

2013-03-22 16:47:45

Last modified on

2013-03-22 16:47:45

Owner

Karid (16341)

Last modified by

Karid (16341)

Numerical id

10

Author

Karid (16341)

Entry type

Definition

Classification

msc 20B99

Classification

msc 05A05

Generated on Fri Feb 9 19:21:33 2018 by

LaTeXML
group ring

group ring

For any group

G

, the

group ring

ℤ

⁢

[

G

]

is defined to be the ring whose

additive group

is the

abelian group

of formal

integer

linear combinations

of elements of

G

, and whose

multiplication

operation is defined by multiplication in

G

, extended

ℤ

–linearly to

ℤ

⁢

[

G

]

.

More generally, for any ring

R

, the

group ring

of

G

over

R

is the ring

R

⁢

[

G

]

whose additive group is the abelian group of formal

R

–linear combinations of elements of

G

, i.e.:

R

⁢

[

G

]

:=

{

∑

i

=

1

n

r

i

⁢

g

i

|

r

i

∈

R

,

g

i

∈

G

}

,

and whose multiplication operation is defined by

R

–linearly extending the group multiplication operation of

G

. In the case where

K

is a field, the group ring

K

⁢

[

G

]

is usually called a

group algebra

.

Title

group ring

Canonical name

GroupRing

Date of creation

2013-03-22 12:13:27

Last modified on

2013-03-22 12:13:27

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Definition

Classification

msc 20C05

Classification

msc 20C07

Classification

msc 16S34

Defines

group algebra

Generated on Fri Feb 9 19:21:34 2018 by

LaTeXML
commensurable subgroups

commensurable subgroups

0.1

Definition

Definition -

Let

G

be a group. Two

subgroups

S

1

,

S

2

⊆

G

are said to be

commensurable

, in which case we write

S

1

∼

S

2

, if

S

1

∩

S

2

has finite index both in

S

1

and in

S

2

, i.e. if

[

S

1

:

S

1

∩

S

2

]

and

[

S

2

:

S

1

∩

S

2

]

are both finite.

This can be interpreted informally in the following :

S

1

and

S

2

are commensurable if their

intersection

S

1

∩

S

2

is “big” in both

S

1

and

S

2

.

0.2

Commensurability is an equivalence relation

-

of subgroups is an

equivalence relation

. In particular, if

S

1

∼

S

2

and

S

2

∼

S

3

, then

S

1

∼

S

3

.

:

Let

S

1

,

S

2

and

S

3

be subgroups of a group

G

.

•

Reflexivity

: we have that

S

1

∼

S

1

, since

[

S

1

:

S

1

]

=

1

.

•

Symmetry

: is clear from the definition.

•

Transitivity

: if

S

1

∼

S

2

and

S

2

∼

S

3

, then one has

[

S

1

:

S

1

∩

S

3

]

≤

[

S

1

:

S

1

∩

S

2

∩

S

3

]

=

[

S

1

:

S

1

∩

S

2

]

[

S

1

∩

S

2

:

S

1

∩

S

2

∩

S

3

]

≤

[

S

1

:

S

1

∩

S

2

]

[

S

2

:

S

2

∩

S

3

]

<

∞

.

Similarly, we can prove that

[

S

3

:

S

1

∩

S

3

]

<

∞

and therefore

S

1

∼

S

3

.

□

0.3

Examples:

•

All non-zero subgroups of

ℤ

are commensurable with each other.

•

All

conjugacy classes

of the

general linear group

G

⁢

L

⁢

(

n

;

ℤ

)

, seen as a subgroup of

G

⁢

L

⁢

(

n

;

ℚ

)

, are commensurable with each other.

References

1

A. Krieg,

, Mem. Amer. Math. Soc., no. 435, vol. 87, 1990.

Title

commensurable subgroups

Canonical name

CommensurableSubgroups

Date of creation

2013-03-22 18:34:14

Last modified on

2013-03-22 18:34:14

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

4

Author

asteroid (17536)

Entry type

Definition

Classification

msc 20C08

Related topic

CommensurableNumbers

Defines

commensurable

Generated on Fri Feb 9 19:21:37 2018 by

LaTeXML
a representation which is not completely reducible

a representation which is not completely reducible

If

G

is a

finite group

, and

k

is a field whose

characteristic

does divide the order of the group, then Maschke’s

theorem

fails. For example let

V

be the

regular representation

of

G

, which can be thought of as

functions

from

G

to

k

, with the

G

action

g

⋅

φ

⁢

(

g

′

)

=

φ

⁢

(

g

-

1

⁢

g

′

)

. Then this

representation

is not

completely reducible

.

There is an obvious

trivial

subrepresentation

W

of

V

, consisting of the

constant functions

. I claim that there is no

complementary

invariant subspace

to this one. If

W

′

is such a

subspace

, then there is a

homomorphism

φ

:

V

→

V

/

W

′

≅

k

. Now consider the

characteristic function

of the

identity

e

∈

G

δ

e

⁢

(

g

)

=

{

1

g

=

e

0

g

≠

e

and

ℓ

=

φ

⁢

(

δ

e

)

in

V

/

W

′

. This is not zero since

δ

generates

the representation

V

. By

G

-equivarience,

φ

⁢

(

δ

g

)

=

ℓ

for all

g

∈

G

. Since

η

=

∑

g

∈

G

η

⁢

(

g

)

⁢

δ

g

for all

η

∈

V

,

W

′

=

φ

⁢

(

η

)

=

ℓ

⁢

(

∑

g

∈

G

η

⁢

(

g

)

)

.

Thus,

ker

⁢

φ

=

{

η

∈

V

|

∑

∈

G

η

⁢

(

g

)

=

0

}

.

But since the characteristic of the field

k

divides the order of

G

,

W

≤

W

′

, and thus could not possibly be complementary to it.

For example, if

G

=

C

2

=

{

e

,

f

}

then the invariant subspace of

V

is spanned by

e

+

f

. For characteristics other than

2

,

e

-

f

spans a complementary subspace, but over characteristic 2, these

elements

are the same.

Title

a

representation which is not completely reducible

Canonical name

ARepresentationWhichIsNotCompletelyReducible

Date of creation

2013-03-22 13:31:47

Last modified on

2013-03-22 13:31:47

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

5

Author

bwebste (988)

Entry type

Example

Classification

msc 20C15

Generated on Fri Feb 9 19:21:40 2018 by

LaTeXML
Maschke’s theorem

Maschke’s theorem

Let

G

be a

finite group

, and

k

a field of

characteristic

not dividing

|

G

|

. Then any

representation

V

of

G

over

k

is

completely reducible

.

Proof.

We need only show that any subrepresentation has a

complement

, and the result follows by

induction

.

Let

V

be a representation of

G

and

W

a subrepresentation. Let

π

:

V

→

W

be an arbitrary

projection

, and let

π

′

⁢

(

v

)

=

1

|

G

|

⁢

∑

g

∈

G

g

-

1

⁢

π

⁢

(

g

⁢

v

)

This map is obviously

G

-

equivariant

, and is the

identity

on

W

, and its image is

contained

in

W

, since

W

is

invariant

under

G

. Thus it is an equivariant projection to

W

, and its kernel is a complement to

W

.
∎

Title

Maschke’s

theorem

Canonical name

MaschkesTheorem

Date of creation

2013-03-22 13:21:16

Last modified on

2013-03-22 13:21:16

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

9

Author

bwebste (988)

Entry type

Theorem

Classification

msc 20C15

Generated on Fri Feb 9 19:21:42 2018 by

LaTeXML
immanent

immanent

Let

S

n

denote the

symmetric group

on

n

elements.
Let

χ

:

S

n

→

ℂ

be a complex

character

.
For any

n

×

n

matrix

A

=

(

a

i

⁢

j

)

i

,

j

=

1

n

define the

immanent

of

A

as

Imm

χ

⁡

(

A

)

=

∑

σ

∈

S

n

χ

⁢

(

σ

)

⁢

∏

j

=

1

n

A

j

⁢

σ

⁢

(

j

)

.

Special cases of immanents are determinants and permanents — in the case where

χ

is the constant character (

χ

⁢

(

x

)

=

1

for all

x

∈

S

n

),

Imm

χ

⁡

(

A

)

is the permanent of

A

. In the case where

χ

is the sign of the

permutation

(which is the character of the

permutation group

associated to the (non-trivial) one-dimensional

representation

),

Imm

χ

⁡

(

A

)

is the determinant of

A

.

Title

immanent

Canonical name

Immanent

Date of creation

2013-03-22 14:05:43

Last modified on

2013-03-22 14:05:43

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

17

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 20C30

Related topic

permanent

Related topic

character

Generated on Fri Feb 9 19:21:48 2018 by

LaTeXML
Young’s projection operators

Young’s projection operators

Associated to a

Young tableau

with

n

boxes, we have two elements
of the

group ring

of the

permutation group

on

n

symbols. To
construct the operators, we first construct the antisymmetrizing
operators corresponding to the columns and the symmetrizing
operators corresponding to the rows. Then one operator corresponding
to the tableau consists of the product of the symmetrizing operators
corresponding to the rows multiplied by the product of the
antisymmetrizing operators corresponding to the columns and the
other consists of the product of the antisymmetrizing operators
corresponding to the columns multiplied by the symmetrizing operators
corresponding to the rows.

How this works may be illustrated with a simple example. Consider
the tableau

1

2

3

4

.

Corresponding to the first row, we have the symmetrization operator

1

+

(

1 2

)

.

Corresponding to the second row, we have the symmetrization operator

1

+

(

3 4

)

Multiplying these two symmetrization operators (the order does not
matter because they involve

permutations

of different elements)
produces

1

+

(

1 2

)

+

(

3 4

)

+

(

1 2

)

⁢

(

3 4

)

Corresponding to the first column, we have the antisymmetrization
operator

1

-

(

1 3

)

.

Corresponding to the second column, we have the antisymmetrization
operator

1

-

(

2 4

)

.

Multiplying these two antisymmetrization operators (the order does not
matter because they involve permutations of different elements)
produces

1

-

(

1 3

)

-

(

2 4

)

+

(

1 3

)

⁢

(

2 4

)

.

To obtain one Young projector, we multiply the product of
the symmetrization operators by the product of the
antisymmetrization
operators.

(

1

+

(

1 2

)

+

(

3 4

)

+

(

1 2

)

⁢

(

3 4

)

)

⁢

(

1

-

(

1 3

)

-

(

2 4

)

+

(

1 3

)

⁢

(

2 4

)

)

=

1

+

(

1 2

)

+

(

3 4

)

+

(

1 2

)

⁢

(

3 4

)

-

(

1 3

)

-

(

1 2 3

)

-

(

1 3 4

)

-

(

1 2 3 4

)

-

(

2 4

)

-

(

1 4 2

)

-

(

2 4 3

)

-

(

1 4 3 2

)

+

(

1 3

)

⁢

(

2 4

)

+

(

1 4 2 3

)

+

(

1 3 2 4

)

+

(

1 4

)

⁢

(

2 3

)

To obtain the other projector, we multiply in the other order.

(

1

-

(

1 3

)

-

(

2 4

)

+

(

1 3

)

⁢

(

2 4

)

)

⁢

(

1

+

(

1 2

)

+

(

3 4

)

+

(

1 2

)

⁢

(

3 4

)

)

=

1

+

(

1 2

)

+

(

3 4

)

+

(

1 2

)

⁢

(

3 4

)

-

(

1 3

)

-

(

1 3 2

)

-

(

1 4 3

)

-

(

1 4 3 2

)

-

(

2 4

)

-

(

1 2 4

)

-

(

2 3 4

)

-

(

1 2 3 4

)

+

(

1 3

)

⁢

(

2 4

)

+

(

1 3 2 4

)

+

(

1 4 2 3

)

+

(

1 4

)

⁢

(

2 3

)

Title

Young’s projection operators

Canonical name

YoungsProjectionOperators

Date of creation

2013-03-22 16:48:25

Last modified on

2013-03-22 16:48:25

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

16

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 20C30

Classification

msc 11P99

Classification

msc 05A17

Generated on Fri Feb 9 19:21:51 2018 by

LaTeXML
unitary representation

unitary representation

Let

G

be a

topological group

. A

unitary representation

of

G

is a pair

(

π

,

H

)

where

H

is a

Hilbert space

and

π

:

G

→

U

⁢

(

H

)

is a

homomorphism

such that
the mapping of

G

×

H

→

H

that sends

(

g

,

v

)

to

π

⁢

(

g

)

⁢

v

is

continuous

. Here

U

⁢

(

H

)

denotes the set of

unitary operators

of

H

.
The group

G

is said to act unitarily on

H

or sometimes,

G

is said to act by unitary representation on

H

.

Title

unitary representation

Canonical name

UnitaryRepresentation

Date of creation

2013-03-22 16:51:45

Last modified on

2013-03-22 16:51:45

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

4

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 20C35

Related topic

IrreducibleUnitaryRepresentationsOfCompactGroupsAreFiniteDimensional

Generated on Fri Feb 9 19:21:54 2018 by

LaTeXML
algebra

algebra

In this definition, all rings are assumed to be

rings with identity

and all

ring homomorphisms

are assumed to be unital.

Let

R

be a ring. An

algebra

over

R

is a ring

A

together with a ring homomorphism

f

:

R

→

Z

⁢

(

A

)

, where

Z

⁢

(

A

)

denotes the center of

A

. A

subalgebra

of

A

is a subset of

A

which is an algebra.

Equivalently, an algebra over a ring

R

is an

R

–module

A

which is a ring and satisfies the property

r

⋅

(

x

*

y

)

=

(

r

⋅

x

)

*

y

=

x

*

(

r

⋅

y

)

for all

r

∈

R

and all

x

,

y

∈

A

. Here

⋅

denotes

R

-module multiplication and

*

denotes

ring multiplication

in

A

. One passes between the two definitions as follows: given any ring homomorphism

f

:

R

⟶

Z

⁢

(

A

)

, the

scalar multiplication

rule

r

⋅

b

:=

f

⁢

(

r

)

*

b

makes

A

into an

R

-module in the sense of the second definition. Conversely, if

A

satisfies the requirements of the second definition, then the function

f

:

R

→

A

defined by

f

⁢

(

r

)

:=

r

⋅

1

is a ring homomorphism from

R

into

Z

⁢

(

A

)

.

Title

algebra

Canonical name

Algebra

Date of creation

2013-03-22 11:48:37

Last modified on

2013-03-22 11:48:37

Owner

djao (24)

Last modified by

djao (24)

Numerical id

17

Author

djao (24)

Entry type

Definition

Classification

msc 20C99

Classification

msc 16S99

Classification

msc 13B02

Defines

subalgebra

Generated on Fri Feb 9 19:21:56 2018 by

LaTeXML