 Character - Maple Help

GroupTheory

 Character
 construct a finite group character from a character table Calling Sequence Character( ct, k ) Parameters

 ct - character table k - positive integer Description

 • The Character( ct, k ) command creates the kth character of the character table ct, which may be constructed by using the CharacterTable command from a finite group. Methods

 • Characters are implemented as Maple objects and support several object methods, outlined below.

 Indicator( chi, k ) returns the $k$th higher indicator of the character chi Indicator( chi ) returns the Frobenius-Schur indicator of the character chi Kernel( chi ) returns the kernel of the character chi chi1 . chi2 returns the inner product of characters chi1 and chi2 Examples

 > $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
 > $G≔\mathrm{Alt}\left(4\right)$
 ${\mathrm{GroupTheory}}{:-}{\mathrm{AlternatingGroup}}{}\left({4}\right)$ (1)
 > $\mathrm{ct}≔\mathrm{CharacterTable}\left(G\right)$
 ${\mathrm{GroupTheory}}{:-}{\mathrm{CharacterTable}}{}\left({\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}\right)$ (2)

Retrieve the fourth character (i.e., the last row) from the character table.

 > $\mathrm{Character}\left(\mathrm{ct},4\right)$
 ${\mathrm{GroupTheory}}{:-}{\mathrm{Character}}{}\left({\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{4}\right)$ (3)

The next statement assigns all the characters from the table.

 > $\mathrm{c1},\mathrm{c2},\mathrm{c3},\mathrm{c4}≔\mathrm{op}\left(\mathrm{map2}\left(\mathrm{Character},\mathrm{ct},\left[\mathrm{seq}\right]\left(1..4\right)\right)\right)$
 ${\mathrm{c1}}{,}{\mathrm{c2}}{,}{\mathrm{c3}}{,}{\mathrm{c4}}{≔}⟨{\text{character:}}{\text{1a}}{\to }{1}{,}{\text{2a}}{\to }{1}{,}{\text{3a}}{\to }{1}{,}{\text{3b}}{\to }{1}{\text{for}}{{\mathbf{A}}}_{{4}}⟩{,}⟨{\text{character:}}{\text{1a}}{\to }{1}{,}{\text{2a}}{\to }{1}{,}{\text{3a}}{\to }{-}\frac{{1}}{{2}}{-}\frac{{I}{}\sqrt{{3}}}{{2}}{,}{\text{3b}}{\to }{-}\frac{{1}}{{2}}{+}\frac{{I}{}\sqrt{{3}}}{{2}}{\text{for}}{{\mathbf{A}}}_{{4}}⟩{,}⟨{\text{character:}}{\text{1a}}{\to }{1}{,}{\text{2a}}{\to }{1}{,}{\text{3a}}{\to }{-}\frac{{1}}{{2}}{+}\frac{{I}{}\sqrt{{3}}}{{2}}{,}{\text{3b}}{\to }{-}\frac{{1}}{{2}}{-}\frac{{I}{}\sqrt{{3}}}{{2}}{\text{for}}{{\mathbf{A}}}_{{4}}⟩{,}⟨{\text{character:}}{\text{1a}}{\to }{3}{,}{\text{2a}}{\to }{-1}{,}{\text{3a}}{\to }{0}{,}{\text{3b}}{\to }{0}{\text{for}}{{\mathbf{A}}}_{{4}}⟩$ (4)
 > $\mathrm{c4}·\mathrm{c4}$
 ${1}$ (5)
 > $\mathrm{c3}·\mathrm{c4}$
 ${0}$ (6)

Since the irreducible characters form an orthonormal basis, the following produces an identity matrix.

 > $\mathrm{Matrix}\left(4,4,\left(i,j\right)↦c‖i·c‖j\right)$
 $\left[\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$ (7)
 > $K≔\mathrm{Kernel}\left(\mathrm{c3}\right)$
 ${\mathrm{GroupTheory}}{:-}{\mathrm{PermutationGroup}}{}\left(\left\{{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}\right\}{,}{\mathrm{degree}}{=}{4}{,}{\mathrm{supergroup}}{=}{\mathrm{GroupTheory}}{:-}{\mathrm{PermutationGroup}}{}\left(\left\{{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}\right\}{,}{\mathrm{degree}}{=}{4}\right)\right)$ (8)
 > $\mathrm{IsNormal}\left(K,G\right)$
 ${\mathrm{true}}$ (9)
 > $\mathrm{GetValues}\left(\mathrm{c3}\right)$
 $\left[{1}{,}{1}{,}{-}\frac{{1}}{{2}}{+}\frac{{I}{}\sqrt{{3}}}{{2}}{,}{-}\frac{{1}}{{2}}{-}\frac{{I}{}\sqrt{{3}}}{{2}}\right]$ (10)

Notice that, although the non-Abelian groups of order $8$ have identical character tables, they are distinguished by the Frobenius-Schur indicator.

 > $\mathrm{ctQ}≔\mathrm{CharacterTable}\left(\mathrm{QuaternionGroup}\left(\right)\right):$
 > $\mathrm{c1Q},\mathrm{c2Q},\mathrm{c3Q},\mathrm{c4Q},\mathrm{c5Q}≔\mathrm{op}\left(\mathrm{map2}\left(\mathrm{Character},\mathrm{ctQ},\left[\mathrm{seq}\right]\left(1..5\right)\right)\right):$
 > $\mathrm{ctD4}≔\mathrm{CharacterTable}\left(\mathrm{DihedralGroup}\left(4\right)\right):$
 > $\mathrm{c1D4},\mathrm{c2D4},\mathrm{c3D4},\mathrm{c4D4},\mathrm{c5D4}≔\mathrm{op}\left(\mathrm{map2}\left(\mathrm{Character},\mathrm{ctD4},\left[\mathrm{seq}\right]\left(1..5\right)\right)\right):$
 > $\mathrm{map}\left(\mathrm{Indicator},\left[\mathrm{c1Q},\mathrm{c2Q},\mathrm{c3Q},\mathrm{c4Q},\mathrm{c5Q}\right]\right)$
 $\left[{1}{,}{1}{,}{1}{,}{1}{,}{-1}\right]$ (11)
 > $\mathrm{map}\left(\mathrm{Indicator},\left[\mathrm{c1D4},\mathrm{c2D4},\mathrm{c3D4},\mathrm{c4D4},\mathrm{c5D4}\right]\right)$
 $\left[{1}{,}{1}{,}{1}{,}{1}{,}{1}\right]$ (12)
 > Compatibility

 • The GroupTheory[Character] command was introduced in Maple 2017.