A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components …. There is also the case of an anti-symmetric tensor that is only anti-symmetric in specified pairs of indices. A (or . Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric. Antisymmetric and symmetric tensors Homework Equations The Attempt at a Solution The first bit I think is just like the proof that a symmetric tensor multiplied by an antisymmetric tensor is always equal to zero. Thanks Evgeny, I used Tr(AB T) = Tr(A T B) Tr(A T B)=Tr(AB) and Tr(AB T)=Tr(A(-B))=-Tr(AB) So Tr(AB)=-Tr(AB), therefore Tr(AB)=0 But if it can be done along the lines I tried with indexes, I'd really like to see that - I am looking for opportunities to practice Indexing This makes many vector identities easy to prove. Obviously if something is equivalent to negative itself, it is zero, so for any repeated index value, the element is zero. SOLUTION Since the and are dummy indexes can be interchanged, so that A S = A S = A S = A S 0: Each tensor can be written like the sum of a symmetric part V = 1 2 V + V and an antisymmetric part V~ = 1 2 V V so that a V = V +V~ = 1 2 V +V +V V = V There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. A and B is zero, one says that the tensors are orthogonal, A :B =tr(ATB)=0, A,B orthogonal (1.10.13) 1.10.4 The Norm of a Tensor . I think your teacher means Frobenius product.In the context of tensor analysis (e.g. Using 1.2.8 and 1.10.11, the norm of a second order tensor A, denoted by . Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors . the product of a symmetric tensor times an antisym-metric one is equal to zero. However, the connection is not a tensor? A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Antisymmetric and symmetric tensors. (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. The alternating tensor can be used to write down the vector equation z = x × y in suffix notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 −x 3y 2, as required.) Antisymmetric Tensor By definition, A µν = −A νµ,so A νµ = L ν αL µ βA αβ = −L ν αL µ βA βα = −L µ βL ν αA βα = −A µν (3) So, antisymmetry is also preserved under Lorentz transformations. S = 0, i.e. I see that if it is symmetric, the second relation is 0, and if antisymmetric, the first first relation is zero, so that you recover the same tensor) A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components [math]U_{ijk\dots}[/math] and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: I agree with the symmetry described of both objects. Thus, the doubly contracted product of a symmetric tensor T with any tensor B equals T doubly contracted with the symmetric part of B, and the doubly contracted product of a symmetric tensor and an antisymmetric tensor is zero. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: A), is Antisymmetric and symmetric tensors. 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