Is the differential d r → a sensible mathematical object?When doing differential geometry, physicists often use d r → = d x i e → i for many different things. For instance, they define the holonomic basis { e → a ′ } relative to a coordinate system { x ′ a } by imposing d r → = d x ′ a e → a ′ ⟹ e → a ′ = ∂ r → ∂ x ′ a and they compute the quadratic form of the metric d s 2 as d r → ⋅ d r → .Computing the differential of a vector field ( r → = x i e → i i, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating k-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).Not only that, the "true" definitions of holonomic bases and ds2 don't use this d r → at all.EDIT: in fact, taking the derivative of r → , or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn't define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors ∂ / ∂ x ′ a .After thinking about it, I thought the differential of a vector field might just be d φ → = ( ∇ i φ j ) e → j ⊗ d x i ,so maybe d r → = d x i e → i i means d r → = d x i ⊗ e → i ? How is d r → rigorously defined, otherwise?

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Is the differential       d              r      →       a sensible mathematical object?When doing differential geometry, physicists often use      d              r      →        =      d        x    i                         e        →              i  for many different things. For instance, they define the holonomic basis   {                    e        →              a                 ′        } relative to a coordinate system   {      x          ′      a        } by imposing      d              r      →        =      d        x          ′              a                                   e        →              a                 ′          ⟹                      e        →              a                 ′        =            ∂                        r          →                            ∂              x                  ′          a                    and they compute the quadratic form of the metric       d        s    2   as       d              r      →        ⋅      d              r      →       .Computing the differential of a vector field (            r      →        =      x    i                      e        →              i   i, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating k-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).Not only that, the &quot;true&quot; definitions of holonomic bases and ds2 don&#039;t use this       d              r      →       at all.EDIT: in fact, taking the derivative of             r      →      , or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn&#039;t define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors   ∂      /    ∂      x          ′      a      .After thinking about it, I thought the differential of a vector field might just be      d              φ      →        =  (      ∇    i        φ    j    )                       e        →              j    ⊗      d        x    i    ,so maybe       d              r      →        =      d        x    i                         e        →              i   i means       d              r      →        =      d        x    i    ⊗                    e        →              i  ? How is       d              r      →       rigorously defined, otherwise?

Adolfo Lee · Accepted Answer

You&#039;re correct. it&#039;s a tensor product. Abstractly, on a vector space V,   d            r      →        ∈  V  ⊗      V    ∗  . Any element of this tensor space defines a map   V  →  V. For example, if   e  ⊗  ω  ∈  V  ⊗      V    ∗  , then it defines a map  v  ↦  ⟨  ω  ,  v  ⟩  e  .In particular, given any frame of vector fields,   (                    e        →              1    ,  …  ,                    e        →              n    ) with the dual basis of 1-form   (      ω    1    ,  …  ,      ω    n    ), the definition of   d            r      →       is  d            r      →        =                    e        →              i    ⊗      ω    i  This definition is invariant under change of basis. In particular, the map associated with it is simply the identity map,  ⟨  d            r      →        ,  v  ⟩  =  ⟨                    e        →              i    ⊗      ω    i    ,      v    j        e    j    ⟩  =                    e        →              i        v    j    ⟨      ω    i    ,      e    j    ⟩  =      v    i        e    i    =  v  .If you have a coordinate system   (      x    1    ,  …  ,      x    n    ), then you can set   (                    e        →              1    ,  …  ,                    e        →              n    ) equal to the coordinate vector fields and   (      ω    1    ,  …  ,      ω    n    )  =  (  d      x    1    ,  …  ,  d      x    n    ).If you have another coordinate system   (      y    1    ,  …  ,      y    n    ) with corresponding coordinate vector fields   (                    f        →              1    ,  …  ,                    f        →              n    ) and dual frame   (  d      y    1    ,  …  ,  d      y    n    ), then  d            r      →        =                    e        →              i    ⊗  d      x    i    =                    f        →              i    ⊗  d      y    i    .From here, I think it should be straightforward to derive all of the formulas you&#039;ve written down.

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