How Do Tensor and Wedge Products Relate in Differential Geometry?

[mathwonk]

i am a little puzzled no one who went through bachmans book seemed to notice this discrepancy.

probably ordinary tensors were never considered there.

 

[mathwonk]

lets see. if we consider the operation of interchanging entrioes in a 2 tensor, i.e. taking atensb to btens a, we geta linear endomorphism of the space of 2 tensors that is an involution, i.e. satisfies T^2 = I or T^2-I = 0. thus its minimal polynomial factors as (T-I)(T+I), and we hVE EIGENVALUES 1 and -1.

hence the space should decompose into eigenspaces of T-I and T+I, namely symmetric and antisymmetric tensors.

so this seems to be why every 2 tensor decomposes this way, from the point of view of spectral theory.but what to saY ABoUT 3 tensors??

 

[cathalcummins]

If we were to generalise this to general 3-forms. \omega \in T^{0}_{3} such that T_{ijk}=-T_{jik}. The thing I am trying to prove is that n-forms form a vector space, and to find the dimension of this space. To make generalisation easier, could you please clarify the following reasoning.

Would it be correct to split T^{0}_{3} into its symmetric and anti symmetric parts, S and A such that;

S_{ijk}=\frac{1}{2}(T_{ijk}+T_{jik})

and

A_{ijk}=\frac{1}{2}(T_{ijk}-T_{jik})

Then, as \omega[/tex] is by definition anti-symmetric, its coefficients must have the form of the A&#039;s. <br /> <br /> 1) The dimension would be the number of independent components of A yes? for the 2 form, this is easily seen from the upper triangular (not including the diagonal), which is just n(n-1)/2. My problem is trying to generalise three forms and upwards. Mainly because I can visualise the permutations very well.<br /> <br /> Thanks