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Outline

An Easy Construction of G2

2008

Abstract

Abstract. We show how to construct the simple exceptional Lie algebra of type G2 by explicitly constructing its 7 dimensional representation. No knowledge of Lie theory is required, and all relevant coefficients can be remembered by use of the simple mnemonic βαββαβ. 1.

FAQs

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What unique methods facilitate the construction of the G2 Lie algebra?add

The paper demonstrates a novel construction using the directed G2 Hexagon graph combined with mutations, yielding an easier approach compared to traditional methods. Specifically, it requires no prior knowledge of Lie theory, significantly simplifying the process.

How do mutations and graphs relate to G2's structure constants?add

The study shows that mutations in directed graphs help derive weights and structure constants for G2. The explicit commutation relations can be traced back to observed relationships during mutations across the G2 Hexagon.

What is the dimensionality of the Lie algebra g2 constructed in this paper?add

The construction reveals that the Lie algebra g2 has a dimension of 14, originating from the operators defined on the graph's vertices. Each operator corresponds to vertices in the G2 Hexagon, facilitating easier computations of relations.

What role do dual mutations play in the Numbers Game for G2?add

Dual mutations in the Numbers Game adaptively change vertex functions, leading to a symmetric structure of roots. This process generates positive roots reflecting key properties of the algebra and aids in the construction of the G2 framework.

How does the construction differ from traditional Lie theory approaches?add

Unlike traditional constructions reliant on complex Lie theory, this paper uses graph theory and explicit operator definitions. This fresh perspective not only simplifies understanding but also broadens the possible applications within representation theory.

References (18)

  1. X β , X α ] = X α+β [X α+β , X β ] = 2X α+2β [X α+2β , X β ] = 3X α+3β [X α+3β , X α ] = X 2α+3β [X α+2β , X α+β ] = 3X 2α+3β [Y β , Y α ] = Y α+β [Y α+β , Y β ] = 2Y α+2β [Y α+2β , Y β ] = 3Y α+3β [Y α+3β , Y α ] = Y 2α+3β [Y α+2β , Y α+β ] = 3Y 2α+3β [X α , Y α+β ] = -Y β [X α , Y 2α+3β ] = -Y α+3β [X β , Y α+3β ] = -Y α+2β [X β , Y α+2β ] = -2Y α+β [X β , Y α+β ] = 3Y α [X α+β , Y α = X β [X α+β , Y β ] = -3X α [X α+β , Y α+2β ] = -2Y β [X α+2β , Y β ] = 2X α+β [X α+2β , Y α+β ] = 2X β [X α+2β , Y α+3β = Y α [X α+2β , Y 2α+3β = Y α+β [X α+3β , Y 2α+3β ] = -Y α [X 2α+3β , Y α ] = X α+3β [X 2α+3β , Y α+β ] = -X α+2β
  2. X 2α+3β , Y α+2β ] = -X α+β [X 2α+3β , Y α+3β ] = X α [Y β , X β ] = H β [Y α , X α ] = H α [Y α+β , X α+β ] = H α+β = 3H α + H β [Y α+2β , X α+2β ] = H α+2β = 2H α + 3H β [Y α+3β , X α+3β ] = H α+3β = H α + H β [Y 2α+3β , X 2α+3β ] = H 2α+3β = 2H α + H β [X γ , H β ] = H β (γ)X γ [Y γ , H β ] = -H β (γ)Y γ [X γ , H α ] = H α (γ)X γ [Y γ , H α ] = -H α (γ)Y γ . example? If so, what is the References
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