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Author Topic: Rotation Group and (0, 2) Tensor  (Read 24109 times)
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« on: February 14, 2007, 06:21:43 PM »

An arbitrary (0, 2) tensor  X_{\mu \nu} can be decomposed into symmetric and antisymmetric parts, with its symmetric parts further decomposed into trace and trace-free parts. It is said in my GR book that in group theory language, such decomposition means that we're looking for "irreducible representations" of the rotation group, i.e. each individual parts transform only into themselves under rotations, and we say that the parts define "invariant subspaces" of the space of (0, 2) tensors.

The problem is I can't see how such composition is linked to representations of the rotation group. Please help. icon adore
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« Reply #1 on: February 14, 2007, 11:06:13 PM »

An arbitrary (0, 2) tensor  X_{\mu \nu} ...

Could you be more specific?  What is the transformation of this tensor? Which space is it in?
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« Reply #2 on: February 14, 2007, 11:37:05 PM »

 X_{\mu \nu} is in the space of (0, 2) tensors. Its transformation is

 \displaystyle X_{\mu^\prime \nu^\prime} = \frac{\partial x^\mu}{\partial x^{\mu^\prime}} \frac{\partial x^\nu}{\partial x^{\nu^\prime}} X_{\mu \nu}

And the aforementioned decomposition is

 X_{\mu \nu} = \frac{1}{n} Xg_{\mu \nu} + \hat{X}_{\mu \nu} + X_{[\mu \nu]}

where the first two terms are the symmetric part (trace + trace-free parts) and the last term is the antisymmetric part. Each term is separately invariant under rotations  \in \mathrm{SO(3, 1)}.

I've attached the page. It talks about just the  ij components and spatial rotations. Using the above equation,  s_{ij} can be found. However, I don't understand the parenthesised note. Why is "decomposing the tensors into such parts" equal to "looking for irreducible representations of the rotation group"?


* degreesoffreedom.gif (83.68 KB, 570x850 - viewed 493 times.)
« Last Edit: February 15, 2007, 12:21:52 AM by เกียรติศักดิ์ » Logged

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« Reply #3 on: February 15, 2007, 08:17:27 AM »

...
And the aforementioned decomposition is

 X_{\mu \nu} = \frac{1}{n} Xg_{\mu \nu} + \hat{X}_{\mu \nu} + X_{[\mu \nu]}

...

I think I can guess that X_{[\mu \nu]} \equiv X_{\mu \nu} - X_{\nu \mu}, but what are the other two?
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« Reply #4 on: February 16, 2007, 01:17:09 PM »

I use this convention instead:

 X_{[\mu \nu]} \equiv \dfrac{1}{N!} \left( X_{\mu \nu} - X_{\nu \mu} \right)

where  N is the number of permuted indices ( N = 2 in our case). Similarly,

 X_{(\mu \nu)} = \dfrac{1}{N!} \left( X_{\mu \nu} + X_{\nu \mu} \right)

So,

 X_{\mu \nu} = X_{(\mu \nu)} + X_{[\mu \nu]}

The antisymmetric (0, 2) tensor is automatically traceless. But the symmetric tensor has both the trace and trace-free parts:

 X_{(\mu \nu)} = \dfrac{1}{n} X g_{\mu \nu} + \hat{X}_{\mu \nu}

where  X is the trace and  \hat{X}_{\mu \nu} is a traceless symmetric tensor.
« Last Edit: February 16, 2007, 01:22:08 PM by เกียรติศักดิ์ » Logged

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« Reply #5 on: February 16, 2007, 02:09:15 PM »

Each term is separately invariant under rotations  \in \mathrm{SO(3, 1)}.

By this I mean that for  \dfrac{\partial x^\mu}{\partial x^{\mu^\prime}} \in \mathrm{SO(3, 1)}

 \dfrac{\partial x^\mu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\nu}{\partial x^{\nu^\prime}} (X g_{\mu \nu}) = X g_{\mu^\prime \nu^\prime}

i.e. the trace part is still purely the trace part after rotations.

 \hat{X}_{(\mu^\prime \nu^\prime)} = \dfrac{\partial x^\mu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\nu}{\partial x^{\nu^\prime}} \hat{X}_{(\mu \nu)} = \dfrac{\partial x^\nu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\mu}{\partial x^{\nu^\prime}} \hat{X}_{(\nu \mu)} = \dfrac{\partial x^\nu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\mu}{\partial x^{\nu^\prime}} \hat{X}_{(\mu \nu)} = \hat{X}_{(\nu^\prime \mu^\prime)}

i.e. the (traceless) symmetric part is still symmetric after rotations, and similarly

 X_{[\mu^\prime \nu^\prime]} = \dfrac{\partial x^\mu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\nu}{\partial x^{\nu^\prime}} X_{[\mu \nu]} = \dfrac{\partial x^\nu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\mu}{\partial x^{\nu^\prime}} X_{[\nu \mu]} = -\dfrac{\partial x^\nu}{\partial x^{\mu^\prime}} \dfrac{\partial x^\mu}{\partial x^{\nu^\prime}} X_{[\mu \nu]} = -X_{[\nu^\prime \mu^\prime]}

i.e. the antisymmetric part is still antisymmetric after rotations.

Now let's get back to the question: Why is "decomposing the tensors into such parts" equal to "looking for irreducible representations of the rotation group"?
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« Reply #6 on: February 16, 2007, 02:15:49 PM »

...
where  X is the trace and  \hat{X}_{\mu \nu} is a traceless symmetric tensor.

How do you define the trace of  {X}_{\mu \nu} ?
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« Reply #7 on: February 16, 2007, 02:26:03 PM »

 X = X^\lambda_{\phantom{\lambda} \lambda} = g^{\mu \nu} X_{\mu \nu}

That is, the trace of  X_{\mu \nu} is the sum of the diagonal elements of  X^\mu_{\phantom{\mu} \nu} .
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« Reply #8 on: February 16, 2007, 02:52:46 PM »

Each term is separately invariant under rotations  \in \mathrm{SO(3, 1)}.

...
i.e. the trace part is still purely the trace part after rotations.

...

i.e. the (traceless) symmetric part is still symmetric after rotations, and similarly

...
i.e. the antisymmetric part is still antisymmetric after rotations.

Now let's get back to the question: Why is "decomposing the tensors into such parts" equal to "looking for irreducible representations of the rotation group"?

So I think you have already answered part of your question. Each part that you mentioned above transform among themselves (whatever is antisymmetric remains antisymmetric after transformations, etc. . So each must form a representation. What remains is to show that they are irreducible.  coolsmiley
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« Reply #9 on: February 17, 2007, 01:13:48 AM »

I see! Thank you. smitten

The set of  X g_{\mu \nu} is reducible since the matrix form of  g_{\mu \nu} is diagonalizable, and there's no other matrix involved. So all elements can be brought into block-diagonal form by some similarity transformation. All  X then form a one-dimensional irreducible representation.

But for the other two... representations of a continuous group... oh how irreducibility can be determined... buck2
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« Reply #10 on: February 17, 2007, 09:29:36 PM »

I found this sentence on page 183 of the book "Continuous Groups of Transformations" by Luther Pfahler Eisenhart (Dover, 1961). (You also have it in DjVu format.)

"When the representation is transitive, it is said to be irreducible, in the sense that there is no subspace invariant under all the transformations of the representation."

What does he mean by "transitive representation"?
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« Reply #11 on: February 18, 2007, 09:07:05 AM »

...
But for the other two... representations of a continuous group... oh how irreducibility can be determined... buck2

For general linear group GL(n) the matrix elements \Lambda_{\mu \nu} of a transformation matrix are not subject to any restrictions; so the only process of reduction of X_{\mu \nu} is the symmetization of the indices \mu \nu. Therefore the symmetric and antisymmetric combinations of X_{\mu \nu} are irreducible representation of GL(n). But for certain subgroups of GL(n), a further reduction is possible because of further properties. I think that for SO(3,1) you must have something like \Lambda^{\text{T}}g\Lambda=g. This will give more operation that you can play with the indices \mu \nu, i.e. the contration of the indices. The extra condition will make the traceless tensors tranform among themselves and form subspaces which are invariant under SO(3,1). No further reduction is possible at this stage because there is no further properties you can use. Therefore all the representations you get after the processes of symmetrization and contraction are irreducible.  coolsmiley  Correct me if I am wrong.  Grin
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« Reply #12 on: February 18, 2007, 10:09:11 AM »

I found this sentence on page 183 of the book "Continuous Groups of Transformations" by Luther Pfahler Eisenhart (Dover, 1961). (You also have it in DjVu format.)

"When the representation is transitive, it is said to be irreducible, in the sense that there is no subspace invariant under all the transformations of the representation."

What does he mean by "transitive representation"?

I cannot find the book.  Sad

Do you want the meaning of transitive or representation?
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« Reply #13 on: February 18, 2007, 02:02:28 PM »

Yes yes yes... thinking about what we can play with the indices, if there're no more, then that's it. smitten

I think the ability to consider additional operation like contraction does not arise from \Lambda^{\text{T}}g\Lambda=g and the fact that we are talking about  \mathrm{SO} (3, 1) , but because our present representation comprises not ordinary matrices, but ones that portray (0, 2) tensors. So all the properties we can use to reduce them are those of such tensors; contraction can always be done. The problem is what property the contraction does make use of.

Let's think about what contraction does. Since it is only permissible to contract an upper index with a lower one (otherwise the result is not a tensor), one must first raise one index:

 X_{\mu \nu} = g_{\mu \rho} X^\rho_{\phantom{\rho} \nu} \qquad \Rightarrow \qquad g^{\rho \mu} X_{\mu \nu} = X^\rho_{\phantom{\rho} \nu}

then contract them:

 X^\rho_{\phantom{\rho} \nu} \qquad \Rightarrow \qquad X = X^\rho_{\phantom{\rho} \rho}

The result is then called a trace as it's just the sum of diagonal elements of the matrix representing that (1, 1) tensor. The story continues. Consider

 X^\rho_{\phantom{\rho} \rho} = g^{\rho \mu} X_{\mu \rho}

since  g^{\rho \mu} is symmetric, only the symmetrized part  X_{(\mu \rho)} contributes. In principle, if  X \neq 0 in the first place, with clever guess or whatever, we can write  X_{(\mu \rho)} as

 \\ X_{(\mu \rho)} = \begin{pmatrix} x_1 & \cdot & \cdot & \cdot \\ \cdot & x_2 & \cdot & \cdot \\ \cdot & \cdot & x_3 & \cdot \\ \cdot & \cdot & \cdot & x_4 \end{pmatrix} = \begin{pmatrix} y_1 & \circ & \circ & \circ \\ \circ & y_2 & \circ & \circ \\ \circ & \circ & y_3 & \circ \\ \circ & \circ & \circ & y_4 \end{pmatrix} + \begin{pmatrix} z_1 & \bullet & \bullet & \bullet \\ \bullet & z_2 & \bullet & \bullet \\ \bullet & \bullet & z_3 & \bullet \\ \bullet & \bullet & \bullet & z_4 \end{pmatrix} \\ \phantom{X_{(\mu \rho)}} = \widetilde{X}_{\mu \rho} + \widehat{X}_{\mu \rho}

such that the first matrix has a non-zero trace ( \widetilde{X} = X ) which is in general not zero, and the second one is traceless. Because their traces are scalars, For easiness,  \widetilde{X}_{\mu \rho} is written as  \dfrac{1}{n} X g_{\mu \rho} such that  g^{\mu \rho} \widetilde{X}_{\mu \rho} = \dfrac{1}{n} X g^{\mu \rho} g_{\mu \rho} = \dfrac{1}{n} X \delta^\lambda_\lambda = X ; as can be seen above, under rotations, "traceful" tensors transform into traceful tensors, and traceless tensors transform into traceless ones, regardless of whether the rotations are from  \mathrm{SO} (n) or  \mathrm{SO} (n, 1) .

Am I correct? Smiley
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« Reply #14 on: February 18, 2007, 02:09:58 PM »

Do you want the meaning of transitive or representation?

I know what a transitive group is, but I don't know what a transitive representation is. buck2
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