Character determines representation in characteristic zero - Revision history

Vipul: /* Proof */

2011-07-21T14:07:57Z

Proof

← Older revision		Revision as of 14:07, 21 July 2011
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	\| 2 \|\|<math>a_{1i} = \frac{1}{m}\langle \chi,\chi_i \rangle_G</math> and <math>a_{2i} = \frac{1}{m} \langle \chi, \chi_i \rangle_G</math> for each <math>1 \le i \le s</math> \|\| \|\| <math>K</math> has characteristic zero, so the manipulation makes sense \|\| Step (1) \|\|		\| 2 \|\|<math>a_{1i} = \frac{1}{m}\langle \chi,\chi_i \rangle_G</math> and <math>a_{2i} = \frac{1}{m} \langle \chi, \chi_i \rangle_G</math> for each <math>1 \le i \le s</math> \|\| \|\| <math>K</math> has characteristic zero, so the manipulation makes sense \|\| Step (1) \|\|
	\|-		\|-
	\| 3 \|\| <math>a_{1i} = a_{2i}</math> for each <math>1 \le i \le s</math> \|\| \|\| <math>K</math> has characteristic zero \|\| \|\| <toggledisplay>Note that if <math>K</math> had characteristic <math>p</math>, we could only conclude equality modulo <math>p</math>, and not equality as nonnegative integers.</toggledisplay>		\| 3 \|\| <math>a_{1i} = a_{2i}</math> for each <math>1 \le i \le s</math> \|\| \|\| <math>K</math> has characteristic zero \|\| Step (2) \|\| <toggledisplay>Note that if <math>K</math> had characteristic <math>p</math>, we could only conclude equality modulo <math>p</math>, and not equality as nonnegative integers.</toggledisplay>
	\|-		\|-
	\| 4 \|\| <math>\rho_1</math> and <math>\rho_2</math> are equivalent \|\| \|\| \|\| Step (3) \|\|		\| 4 \|\| <math>\rho_1</math> and <math>\rho_2</math> are equivalent \|\| \|\| \|\| Step (3) \|\|
	\|}		\|}

Vipul: /* Related facts */

2011-07-21T14:07:27Z

Related facts

← Older revision		Revision as of 14:07, 21 July 2011
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	==Related facts==		==Related facts==

			===Opposite facts===

	* [[Character does not determine representation in any prime characteristic]]: The problem is that we can construct representations whose character is identically zero simply by adding <math>p</math> copies of an irreducible representation to itself.		* [[Character does not determine representation in any prime characteristic]]: The problem is that we can construct representations whose character is identically zero simply by adding <math>p</math> copies of an irreducible representation to itself.

			===Applications===

			* [[Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero]]

	==Facts used==		==Facts used==

Vipul: /* Proof */

2011-07-13T18:35:11Z

Proof

← Older revision		Revision as of 18:35, 13 July 2011
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	and		and

	<math>\rho_2 \cong a_{21}\varphi_1 \oplus a_{22}\varphi_2 \oplus \dots \oplus a_{1s}\varphi_s</math>		<math>\rho_2 \cong a_{21}\varphi_1 \oplus a_{22}\varphi_2 \oplus \dots \oplus a_{2s}\varphi_s</math>

	Let <math>\chi_i</math> denote the character of <math>\varphi_i</math> and denote by <math>m_i</math> the value <math>\langle \chi_i, \chi_i\rangle_G</math> (note: this would be 1 if <math>K</math> were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).		Let <math>\chi_i</math> denote the character of <math>\varphi_i</math> and denote by <math>m_i</math> the value <math>\langle \chi_i, \chi_i\rangle_G</math> (note: this would be 1 if <math>K</math> were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).

Vipul at 18:34, 13 July 2011

2011-07-13T18:34:48Z

@@ Line 2: / Line 2: @@
 Suppose <math>G</math> is a [[finite group]] and <math>K</math> is a [[field]] of characteristic zero. Then, the [[character]] of any finite-dimensional representation of <math>G</math> over <math>K</math> completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.
 ==Related facts==
@@ Line 10: / Line 12: @@
 # [[uses::Character orthogonality theorem]]
 # [[uses::Orthogonal projection formula]]

Vipul at 03:36, 13 July 2011

2011-07-13T03:36:51Z

← Older revision		Revision as of 03:36, 13 July 2011
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	==Facts used==		==Facts used==

			# [[uses::Character orthogonality theorem]]
	# [[uses::Orthogonal projection formula]]		# [[uses::Orthogonal projection formula]]

Vipul: Created page with "==Statement== Suppose $G$ is a finite group and $K$ is a field of characteristic zero. Then, the character of any finite-dimensional representa..."

2011-07-13T01:59:53Z

Created page with "==Statement== Suppose <math>G</math> is a finite group and <math>K</math> is a field of characteristic zero. Then, the character of any finite-dimensional representa..."

New page

==Statement==

Suppose <math>G</math> is a [[finite group]] and <math>K</math> is a [[field]] of characteristic zero. Then, the [[character]] of any finite-dimensional representation of <math>G</math> over <math>K</math> completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.

==Related facts==

* [[Character does not determine representation in any prime characteristic]]: The problem is that we can construct representations whose character is identically zero simply by adding <math>p</math> copies of an irreducible representation to itself.

==Facts used==

# [[uses::Orthogonal projection formula]]

Character determines representation in characteristic zero - Revision history

Vipul: /* Proof */

Vipul: /* Related facts */

Vipul: /* Proof */

Vipul at 18:34, 13 July 2011

Vipul at 03:36, 13 July 2011

Vipul: Created page with "==Statement== Suppose G is a finite group and K is a field of characteristic zero. Then, the character of any finite-dimensional representa..."

Vipul: Created page with "==Statement== Suppose $G$ is a finite group and $K$ is a field of characteristic zero. Then, the character of any finite-dimensional representa..."