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Exterior product of direct sum

  1. Feb 3, 2010 #1

    I am edging my way towards Dolbeault cohomology on a complex manifold and one of the constructions involves taking the kth exterior product of a direct sum (the decomposition of the cotangent bundle into holomorphic and antiholomorphic subspaces). This relies on a theorem from multilinear algebra that says that the result is the direct sum of tensor products of exterior products of the subspaces (sorry but I do not have the Latex to set down the formula).

    My problem is that when coordinates are subsequently used, the tensor product metamorphoses into a wedge product and I cannot convince myself that these are equivalent. I have not found a proof of the multilinear algebra theorem so I am not clear why a tensor product is required in the first place.

    I appreciate that this is a rather detailed question of technique but it won't leave me alone!

    Any suggestions?

    Thank you in anticipation
  2. jcsd
  3. Feb 4, 2010 #2
  4. Feb 5, 2010 #3
    Thank you for your response.
    The reference that you give is indeed the eventual result required, but my query relates to the algebraic establishment of this via the general result (e.g. Wikipedia, "Exterior Algebra" section 4.2 on Direct Sums which starts
    "In particular, the exterior algebra of a direct sum is isomorphic to the tensor product of the exterior algebras:" ).
    This can of course be applied to the decomposition of the cotangent bundle induced by the almost complex structure. The step that I am missing is the change from a tensor product to a wedge product in the final result that you quote involving local coordinates.

    BTW I have no doubt about the validity of the result that you reference (which can probably be established quite easily by induction) just this underying bit of multilinear algebra.
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