Tensor and wedge products

[cristo]

Ok, one more diff geometry question! I need to show the following:

Show that if C_ij are components of an antisymmetric tensor then $C_{ij}dx^i\otimes dx^j$ corresponds to the 2-form $C_{ij}dx^i\wedge dx^j$.

Now, I know that the wedge product is anti symmetric, and that it can be expressed in terms of the tensor product as follows: $$C_{ij}dx^i\wedge dx^j=C_{ij}\frac{1}{2}(dx^i\otimes dx^j-dx^j\otimes dx^i)$$

However, I can't for the life of me show what is required!

Any hints would be much appreciated!

 

[Hurkyl]

Have you invoked the antisymmetry of C?

 

[cristo]

Have you invoked the antisymmetry of C?

Well, yes, I did try that, but on invoking the antisymmetry of C, we obtain

$$C_{ij}dx^i\wedge dx^j=\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i)$$

And I can't seem to progress from there.

 

[cliowa]

Well, yes, I did try that, but on invoking the antisymmetry of C, we obtain

$$C_{ij}dx^i\wedge dx^j=\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i)$$

And I can't seem to progress from there.

Well, but
$$\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i)=C_{ij}dx^i\otimes dx^j$$

Don't get confused with the indices, just think of what one is doing.

 

[cristo]

Well, but
$$\frac{1}{2}(C_{ij}dx^i\otimes dx^j+C_{ji}dx^j\otimes dx^i)=C_{ij}dx^i\otimes dx^j$$

Don't get confused with the indices, just think of what one is doing.

Haha, I can't believe I didn't see that! Thanks!

 

[mathwonk]

how interesting, you are looking at antisymmetric tensors as if they are a subspace of usual tensors. they can slao be looekd at as a quotient space of usual temsors, then your statement is true for allk tensors if un derstood correctly. i.e. the usual tensor space maps onto the space of alternating tensors by sending dxtensdy to dxwedgedy.

then there also a map back from alternating tensors to usual tensors sending dxwedgedy to (1/2)(dxtensdy - dytensdx).

the composition of these maps is the identity in one direction but not the other.

so i was puzzled by your question, since thinking in terms of the quotient space interpretation, the answer to your question is yes, even for non antisymmetric tensors, and trivially so.

i was confused because you did not define the word "corresponds".

 

[mathwonk]

which in your case meant "equals".

 

[mathwonk]

this an algebra question by the way.

 

[mathwonk]

the algebra of tensros is a little like that of polynomials, as i have tried to explain elsewhere. think about how you write out a homogeneous polynomial, say of degree 2, as a linear combination of x^2, xy, and y^2.

all you need to know is the coefficient of each term if we number the variabkles x,y as say x1 and x2, then we can represent the monomial x^2, i.e. x1x1 just as (1,1), and similarly, we get the representations (1,1),(1,2), (2,2), for x^2, xy, and y^2,

i.e. a symbol a(i,j) is the coeficient of the monomial xixj.

so if we understand summation, then a symbol like a(i,j) means the polynomial which is the sum of the terms a(i,j)xixj.

if we want to do this for non commutative polynomials, we need also xx, xy, yx, yy. so we have coefficients a(i,j) where now there is no way to combine a(i,j) and a(j,i).

here too we can think of usual polynomials as either a quotient or a subspace of non commutative ones.


so we map the non commutaitve polynomials to th usal ones just by sending both xy and yx to xy. well let's make a capital letter for non commutative oens.

so we send XY and YX both to xy.

then we can also map back from reguklar polynomials to non commutaive ones, by sending xy to (1/2)(XY+YX). but obviously xy, where xy = yx, is a simpler object that (1/2)(XY+YX).


this is why it more natural to view anticommutative ones also as a quotient rather than a subspace. but anyway...

spivaks little book calculus on manifolds explains both points of view i believe.

but some physicists learning this stuff are apparently left without any awareness of why tensors are a generalizaton of polynomials.


this insight also makes clear the transformation laws, as all you do to get them is substitute say X1 = (aY1+bY2), and X2 = cY1+dY2, and expand.

again just like polynomials.

 

[mathwonk]

the tensor sign, circle cross, just means the multiplication is not commutative, that's all.

 

[cristo]

Thanks for the insightful comments, mathwonk. I've never come across this explanation of non-commutative polynomials before, and I can see that here, the quotient space is a more natural view. However, for the alternating tensor case, especially since I've only just been introduced to the subject, it seems easier to think of alternating tensors as a subspace of normal tensors, than to picture them as a quotient space. (I realize that this may sound wrong to you; but as a beginner I'm familiar with subspaces, but quotient spaces seem a bit "strange").

Anyway, thanks again for your comments.

 

[mathwonk]

well you are over haflway there when you use the wedge notation, instead of trying to manipulate tensors with antisymmetric coefficients.

the purpose of the wedge notation is to accomplish the same thing as the quotient construction, namely take away the need to deal with expressions like (1/2)(AB-BA), and just use A^B instead.

 

[mathwonk]

i seem to be confused here. does the OP mean to say that

dx^dy corresponds to (1/2)(dxtensdy?.

If so, then this seems odd from one point of view, as it would seem to imply that when acting on the standard unit block e1^e2, the value of dx^dy is 1/2 instead of 1.

? obviously i am more at home with general ideas than computations.

but one thing becoming clear here is that some of these things are merely conventions, and not intrinsic properties.

 

[mathwonk]

so i may be wrong here, but it seems one cannot haVE BOTH THIGNS at once, i.e. the natural correspondence desired by the OP seems to conflict with the volume form dx^dy having the expected value (one) on the standard unit square.

what is the source for your conventions, i.e. what book are you reading?

 

[George Jones]

does the OP mean to say that

dx^dy corresponds to (1/2)(dxtensdy?.

No, this is not what the original post says.

 

[cristo]

i seem to be confused here. does the OP mean to say that

dx^dy corresponds to (1/2)(dxtensdy?.

No, the question in the OP is correct, asking to show that $C_{ij}dx^i\otimes dx^j$ corresponds to $C_{ij}dx^i\wedge dx^j$.

so i may be wrong here, but it seems one cannot haVE BOTH THIGNS at once, i.e. the natural correspondence desired by the OP seems to conflict with the volume form dx^dy having the expected value (one) on the standard unit square.

what is the source for your conventions, i.e. what book are you reading?

The question didn't come from a book, but from the lecture notes for a course on differential geometry that I'm taking. This, he states, is the way to obtain an alternating tensor from two form: $$dx^i\wedge dx^j=\frac{1}{2}(dx^i\otimes dx^j-dx^j\otimes dx^i)$$

 

[mathwonk]

how does saying that dx^dy corresponds to (1/2)(dxtensdy differ from saying that Cij dx^dy corresponds to Cij dxtensdy?

what does it mean to say Cij dx^dy corresponds to Cij dxtensdy? as ia sked, is a summation intended here?

i.e. does this mean say that dx^dy - dy^dx corresponds to

dxtensdy - dytensdx?

if so, then ...OOPs, some of my post seems to have disappeared.

i meant that if he was saying that

dx^dy - dy^dx corresponds to dxtensdy - dytensdx,
then since dx^dy = -dy^dx,

this would imply that 2dx^dy corresponds to dxtensdy - dytensdx,

and this would imply that dx^dy corresponds to
(1/2)(dxtensdy-dytensdx).

but then the value of dx^dy on the unit square (e1,e2)

would equal that of the tensor (1/2)(dxtensdy-dytensdx), namely 1/2.

which seems odd.

 

[mathwonk]

all i am saying is that there sem to be many ways to obtain an alternating tensor from a symbol of form A^B, depending on what one wants to be true.

for the purpose of measuring volumes, i.e. integration, which is hoforms are usually used, the choice above has some curious properties, and seems to differ from the convention in books like spivak, fleming, loomis sternberg, etc...

 

[mathwonk]

there is no hard and fast law about to do it, but there are reasons to ask that

the form dx1^...^dxn act on a sequence of n vectors as the determinant does, i.e. to have dx1^...^dxn correspond to the determinant, not to the determinant divided by (n!).

has your prof defined wedge multiplication between two alternating tensors? if so, how?


i.e. if dx^dy is the wedge product of the two tensors dx and dy, how is it defined?

is dx^dy equal to dxtensdy - dytensdx? or is it (1/2)of that?

 

[cristo]

there is no hard and fast law about to do it, but there are reasons to ask that

the form dx1^...^dxn act on a sequence of n vectors as the determinant does, i.e. to have dx1^...^dxn correspond to the determinant, not to the determinant divided by (n!).

has your prof defined wedge multiplication between two alternating tensors? if so, how?


i.e. if dx^dy is the wedge product of the two tensors dx and dy, how is it defined?

is dx^dy equal to dxtensdy - dytensdx? or is it (1/2)of that?

I remember my prof discussing this. He never defind the wedge of two tensors, but rather defined the wedge of two differential forms: $$(f(x)dx^i\wedge dx^j \wedge \cdots)\wedge(g(x)dx^l\wedge dx^m \wedge \cdots)=f(x)g(x)(dx^i\wedge dx^j \wedge \cdots)\wedge(dx^l \wedge dx^m \wedge \cdots)$$. Then later, when introducing tensors, used the expression $$dx^i\wedge dx^j=1/2(dx^i\otimes dx^j=dx^j\otimes dx^i)$$

He mentioned that the factor 1/2 (1/n! in general) is a convention generally used in mathematical physics texts, whereas in pure maths books the factor is 1.

With respect to the summation; my understanding was that repeated indices were assumed to be summed over unless explicitly stated otherwise.

 

[George Jones]

how does saying that dx^dy corresponds to (1/2)(dxtensdy differ from saying that Cij dx^dy corresponds to Cij dxtensdy?

This is not what the original post says.

is a summation intended here?

Yes, summation over both the repeated indices $i$ and $j$ is implied.

$$\begin{equation*} \begin{split} C&=\sum_{i,j}C_{ij}dx^{i}\otimes dx^{j}\\ &=\frac{1}{2}\sum_{i,j}\left( C_{ij}+C_{ij}\right) dx^{i}\otimes dx^{j}\\ &=\frac{1}{2}\sum_{i,j}\left[ C_{ij}x^{i}\otimes dx^{j}-C_{ji}dx^{i}\otimes dx^{j}\right]\\ &=\frac{1}{2}\sum_{i,j}\left[ C_{ij}dx^{i}\otimes dx^{j}-C_{ij}dx^{j}\otimes dx^{i}\right]\\ &=\sum_{i,j}C_{ij}\frac{1}{2}\left( dx^{i}\otimes dx^{j}-dx^{j}\otimes dx^{i}\right)\\ &=\sum_{i,j}C_{ij}dx^{i}\wedge dx^{j} \end{split} \end{equation*}$$

Here, $C$ is antisymmetric and

$$C_{ij} = C \left( \frac{\partial}{\partial x^i} , \frac{\partial}{\partial x^j} \right).$$

 

[mathwonk]

well now you know the reason for the convention in math books - it makes the form have the right value on the unit cube, namely one.

if you read spivaks calc on manifolds you will see he ahs two definitions, "alternation of a tensor" and wedge product. he defines the wedge product of an r form and an s form, to be (r+s choose r) times the alternation. your professor is defining the wedge product just to equal the alternation.


when you choose a basis for an n dimensional vector space, you automatically get a dual basis for the dual space. there is a one dimensional space of alternating n forms, but two natural ways of specifying a one elment basis, namely one can say that the wedge product of the elements of the dual basis is the basic n form, or one can say that the basic n form is the one with value one on the block spanned by the basic vectors.

the pure math convention sets these two equal to each other. the way your prof is using acts by setting the natural n form to be the one with value (1/n!) on the unit cube, natural from the standpoint of algebra, but a bit odd from the stabndpoint of measure.

this is another reason there is no natural way to view the exterior algebra as inside the tensor algebra. ratehr it is more naturlly viewed qas a quotient, not a subvariety.


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.

 

[mathwonk]

thanks george, and did you see my later posts? the one you read had a typo.

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.

 

[mathwonk]

see page 23 in the book by bachman, at the link in post one of tom mattsons thread, which was widely read and followed here a while back, where the value of dx^dy on the standard square was defined to be one. i,.e,. as a determinant, not 1/2.

by that convention there should still be a 1/2 in the last line of your derivation, no?

 

[mathwonk]

george, your convention that C(ij) = C(ei,ej) seems to imply then that

dx^dy - dy^dx equals one on (e1,e2) which is exactly my point, since then the "area form" dx^dy equals 1/2 on the unit square.

maybe one copuld salvage it by saying that the value assigned to dx^dy is normalized by equaling the value of the area form on the "unit triangle"! (the triangle with sides e1, e2).

no then we get 1/3! for 3 dimensions instead of 1/3. ?

 

[cristo]

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]

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.

 

[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]

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.

 

[mathwonk]

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]

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 [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