# Tensor and wedge products

#### cristo

Staff Emeritus
in fact, over more general rings where one cannot always divide, instead of viewing alternating functions as alternating products of linear functions on the module, one views them as linear functions on the exterior module on the original module.

i.e. instead of dfining alternating functions as ^(V*) where V* is linear functions on V, one defines instead ^V, and then alternating fiunctions on (products of) V is defined as linear functions on ^V, i.e. as (^V)*.

the problem is that in general, ^(V*) is not the same as (^V)* i guess.

gee this is something i would like to know about.
What do you mean by the notation ^(V*)?

V* is the linear functions on V (this is equivalent to the dual space of V, no?). Do you then mean by ^(V*) the alternating functions defined on the dual space V*?

Then for the second case, ^V is defined as the alternating functions on V, and then (^V)* is the dual space of this space of alternating functions?

#### mathwonk

Homework Helper
no, there exists a space called ^r(V) and a universal multiplinear alternating map Vx...xV(r times)--->^r(V), such that the corespondence taking

a linear map ^r(V)--->X to the composition

Vx....xV---->^r(V)--->X, is a one one correpondence between all linear maps out of the wedge, and all multilinear alternating maps out of Vx...xV.

or if you like, in the case of the vector spaces V*, it is just the set of all linear combinations of r fold wedges of elements of V*, i.e. all linear combinations of things like dxi1^...dxir.

this ^r(V*) happens to be isomorphic to linear maps out of ^r(V), i,.e,. to (^rV)*.

i.e. m ultilinear alternating maps out of Vx...xV is naturally isomorphic to (^rV)*, which is a quotient space of the tensor product, and not so naturally isomorphic to the space ^r(V*).

this non naturality is why there is more than one way to set up the correspondence between things like dx^dy and multilinear maps on VxV.

i.e. some people let dx^dy be the map taking e1^e2 to one, others take it to 1/2.

Last edited:

#### Doodle Bob

your derivation still seems to me to have the feature that it makes dx^dy have value 1/2 on the unit square, a somewhat odd choice.
One reason I can think of for the 1/2 convention (which I use btw) is that then, if we break down the (0,2)-tensor $$dx \otimes dy$$ to its skew-symmetric and symmetric parts, then we have $dx \otimes dy= dx \wedge dy + dx \odot dy$, where $\odot$ is the symmetric part.

I like your geometric interpretation of it, though.

#### mathwonk

Homework Helper
yes i noticed that. that was one aspect that i was pursuing as a way of generalizing this stuff to higher order products.

but then it did not seem so clear to me that one could decomposie say a 3 tensor as a sum of alternating and various symmetric parts, so there may be something special about degree 2 here, another sign it is unnatural.

i.e. from that point of view one would "symmetrize" a tensor by adding up all permutations of its indices, and divide by n!.

and one can also antisymmetrize one by adding up all its eprmutations with plus and minus signs according to the sign of the eprmutation.

indeed you seem to need n! different summands.

but what are those other "partially symmetric" components?

but there seems no reason these sums would give back the original tensor as they do in the more trivial case of 2 - tensors.

presumably we would be led into the realm of irreducioble decompositions of group representations, which already makes it interesting, since i knolw zip about that.

but this motivates it for me.

Last edited:

#### mathwonk

Homework Helper
consider for example the antisymnmetrizing construction above sending atensb to some constant times atensb - btensa.

it seems the kernel of this map is the symmetric tensors spanned by those of type atensa.

but look at three terms, atensbtensc, goes to a 6 terms expresion

abc +bca + cab - bac -cba - acb. and the kernel of this construction is not the symmetric tensors at all, but any tensor spanned by ones of form aab.

i.e. then one gets aab + aba + baa - aab - baa - aba.

so this splitting into symmetric plus alternating tensors seems a very peculiar accident of degree 2 tensors.

hence i believe using this 1/2 convention is an ad hoc choice that does not mesh well with the general situation. but as long as you are interested only in 2 tensors it seems fine.

#### mathwonk

Homework Helper
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

Homework Helper
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 [itex]\omega[/tex] is by definition anti-symmetric, its coefficients must have the form of the A's.

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.

Thanks

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving