Maurer-Cartan Form: Is it a One Form?

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SUMMARY

The discussion centers on the classification of the Maurer-Cartan form on a Lie group \( G \) as defined in Nakahara's "Geometry, Topology and Physics." The canonical one-form \( \theta: T_gG \to T_eG \) is debated, with participants asserting that it is indeed a \( \mathfrak{g} \)-valued one-form rather than a traditional one-form. The confusion arises from the tensor product representation \( \theta = V_\mu \otimes \theta^\mu \), leading to the classification of \( \theta \) as a \( (1,1) \) tensor. Ultimately, it is clarified that \( \theta \) retains its identity as a \( \mathfrak{g} \)-valued one-form despite its tensor characteristics.

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  • Understanding of Lie groups and their tangent spaces
  • Familiarity with tensor products and tensor notation
  • Knowledge of differential forms and their properties
  • Basic concepts of vector spaces and linear mappings
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  • Study the properties of \( \mathfrak{g} \)-valued differential forms
  • Explore the relationship between \( (1,1) \) tensors and one-forms in differential geometry
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This discussion is beneficial for mathematicians, physicists, and students specializing in differential geometry, particularly those focusing on Lie groups and their applications in theoretical physics.

Silviu
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Hello! I am reading Geometry Topology and Physics by Nakahara and in Chapter 5.6.4 he defines the canonical (Maurer-Cartan) one form on a Lie group G as: ##\theta : T_gG \to T_eG##. Then he states a theorem in which ##\theta = V_\mu \otimes \theta^\mu##. Both by the tensor product and by the definition from a vector space to another vector space, ##\theta## seems to be a ##(1,1)## tensor, not a one form, as it is stated in the book. Am I missing something? Is it a one form or not? Did I get wrong the definition of one form? Thank you!
 
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It is a ##\mathfrak g## valued one form.
 
Every ##(1,1)## tensor, say ##\varphi \otimes v \in U^* \otimes V## is equivalent to the linear map ##u \longmapsto \varphi(u)\cdot v##. Here we have ##v \in T_gG## and ##\varphi \in T_e^*G##.
 
martinbn said:
It is a ##\mathfrak g## valued one form.
Isn't a one-form by definition a function from vectors to real numbers? And ##\mathfrak{g}## is isomorphic to ##T_gG## so it should behave the same whether it acts on vector fields or on simple vectors (tangents at e). But regardless of how you take it, I still don't understand why there is a basis vector (##V_\mu##), if it acts only on vectors and not on one-forms.
 
fresh_42 said:
Every ##(1,1)## tensor, say ##\varphi \otimes v \in U^* \otimes V## is equivalent to the linear map ##u \longmapsto \varphi(u)\cdot v##. Here we have ##v \in T_gG## and ##\varphi \in T_e^*G##.
I am not sure I understand. Do you mean that a ##(1,1)## tensor can be transformed into a one-form by passing to it a one-form, which cancels the vector basis? But in our case, what would be that one-form? Like to me ##\theta## is clearly a ##(1,1)## tensor, and by the definition in the book the exactly same ##\theta##, without any modification is called a one-form. I am not sure I understand it...
 

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