Using index notation to write d 2 = 0 in terms of a torsion free connection.Let (M,g) be a Riemannian manifold and let ω be a 1-form on M. I want to rewrite d 2 ω = 0 in terms of the Levi-Civita connection.I can show the following: d ω ( X , Y ) = ( ∇ X ω ) ( Y ) − ( ∇ Y ω ) ( X ) ,which in index notation reads ( d ω ) a b = 2 ∇ [ a ω b ] .Similiarly for a 2-form, μ, we have: d μ ( X , Y , Z ) = ( ∇ X μ ) ( Y , Z ) − ( ∇ Y μ ) ( X , Z ) + ( ∇ Z μ ) ( X , Y ) ,which in index notation reads ( d μ ) a b c = 3 ∇ [ a ϕ b c ] .Now plugging in d ω for μ we get 0 = d 2 ω = ( d ( d ω ) ) a b c = ∇ [ a ( d ω ) b c ] .I want to plug in the above expression (in index notation) for dω but I'm not really sure how to handle the indices. Do I just get 3 ∇ [ a 2 ∇ [ b ω c ] ] = 6 ∇ [ a ∇ b ω c ] ?

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Using index notation to write       d    2    =  0  in terms of a torsion free connection.Let (M,g) be a Riemannian manifold and let   ω be a 1-form on M. I want to rewrite       d    2    ω  =  0 in terms of the Levi-Civita connection.I can show the following:  d  ω  (  X  ,  Y  )  =  (      ∇    X    ω  )  (  Y  )  −  (      ∇    Y    ω  )  (  X  )  ,which in index notation reads  (  d  ω      )          a      b        =  2      ∇          [      a            ω          b      ]        .Similiarly for a 2-form,   μ, we have:  d  μ  (  X  ,  Y  ,  Z  )  =  (      ∇    X    μ  )  (  Y  ,  Z  )  −  (      ∇    Y    μ  )  (  X  ,  Z  )  +  (      ∇    Z    μ  )  (  X  ,  Y  )  ,which in index notation reads  (  d  μ      )          a      b      c        =  3      ∇          [      a            ϕ          b      c      ]        .Now plugging in   d  ω for   μ we get  0  =      d    2    ω  =  (  d  (  d  ω  )      )          a      b      c        =      ∇          [      a        (  d  ω      )          b      c      ]        .I want to plug in the above expression (in index notation) for dω but I&#039;m not really sure how to handle the indices. Do I just get  3      ∇          [      a        2      ∇          [      b            ω          c      ]      ]        =  6      ∇          [      a            ∇    b        ω          c      ]        ?

wegpluktee3 · Accepted Answer

Of course, this can be also verified directly by writing out the expansion:                    3                  ∇                      [            a                          2                  ∇                      [            b                                    ω                      c            ]            ]                                              =                  ∇                      a                          2                  ∇                      [            b                                    ω                      c            ]                          +                  ∇                      b                          2                  ∇                      [            c                                    ω                      a            ]                          +                  ∇                      c                          2                  ∇                      [            a                                    ω                      b            ]                                                            =                  ∇                      a                                    ∇                      b                                    ω                      c                          −                  ∇                      a                                    ∇                      c                                    ω                      b                          +                  ∇                      b                                    ∇                      c                                    ω                      a                          −                  ∇                      b                                    ∇                      a                                    ω                      c                          +                  ∇                      c                                    ∇                      a                                    ω                      b                          −                  ∇                      c                                    ∇                      b                                    ω                      a                                                            =        6                  ∇                      [            a                                    ∇          b                          ω                      c            ]                              where we have used that for a tensor tabc with a symmetry      t          a      b      c        =      t          a      [      b      c      ]        =            1      2            (          t              a        b        c              −          t              a        c        b              )  the alternation is expressed by      t          [      a      b      c      ]        =            1      3            (          t              a        b        c              +          t              b        c        a              +          t              c        a        b              )  which is again easy to prove:                              t                      [            a            b            c            ]                                              =                              1            6                                    (                      t                          a              b              c                                −                      t                          a              c              b                                +                      t                          b              c              a                                −                      t                          b              a              c                                +                      t                          c              a              b                                −                      t                          c              b              a                                )                                                  =                              1            6                                    (          2                      t                          a              b              c                                +          2                      t                          b              c              a                                +          2                      t                          c              a              b                                )                    To prove the Poincare lemma       d    2    ω  =  0 locally one can use the Euclidean connection on a chosen coordinate system, and since the exterior derivative is independent of a choice of torsion-free connection, the lemma follows from the fact that partial derivatives commute.

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