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数学

Discovery

Andrius Kulikauskas

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Lietuvių kalba

Introduction E9F5FC

Understandable FFFFFF

Questions FFFFC0

Notes EEEEEE

Software

The four classical root systems are as follows, where throughout, {$i>j$}:

{$A_n$}{$\pm (x_i-x_j)$}
{$B_n$}{$\pm (x_i-x_j), \pm (x_i+x_j), \pm x_i$}
{$C_n$}{$\pm (x_i-x_j), \pm (x_i+x_j), \pm 2x_i$}
{$D_n$}{$\pm (x_i-x_j), \pm (x_i+x_j)$}

We thus see that all classical root systems contain the roots {$\pm (x_i-x_j)$}.

{$C_n$} includes the case {$i=j$}. {$B_n$} and {$D_n$} do not. But {$B_n$} fakes it by having a zero. This yields the same lattice as {$C_n$}. But the root systems bring out the difference. So {$C_n$} gives the conditions needed for this identity. The root systems are building up the context for such an identity.

In particular, {$B_n$}, {$C_n$}, {$D_n$} have two conjugate sets of roots because they are establishing the symmetries underlying a pair of expansions linked by a comparison, an identity. Thus we have a value on the left which is being compared to a value on the right. In the case of {$C_n$} we may have identity, an equal sign. In the case of {$B_n$} we may have a zero. In the case of {$D_n$} we have neither.

The Dynkin diagram for {$D_2$} is a pair of dots. This brings to mind the two conjugate roots of -1. Thus there is this inherent symmetry within {$D_n$}.


Consider this as a problem of the ways of coding counting. And coding is counted by a minus sign. This codes counting both forwards and backwards. But how might the two directions coincide? There are four possibilities. The second direction can be coded with a plus sign. Then the two can be related in three different ways.

IntuitingClassicalRootSystems


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Puslapis paskutinį kartą pakeistas 2018 gruodžio 30 d., 16:54
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