How to derive irreducible representations/tensors?

[Antarres]

I've been using irreducible representations of groups for years now, ever since undergrad studies, but a question that always kind of bothered me, which is maybe a silly question, was to know when I have irreducible representation at hand, and when it isn't the case. Because usually irreducible representation is defined as one that is 'not reducible'. So if I wasn't familiar with the theory of some group etc, where someone has already worked out what irreducible representations/tensors should look like, I would be confused how to derive them myself. In physics it's often about orthogonal/unitary groups which are studied a lot, so you don't really have to pose this question in practice.

Either way, I wanted to make a note for myself when I look at general tensors, how to construct irreducible pieces. And so, instead of looking at a particular group, I wanted to look at general coordinate transformations. Now it is easy to prove that (anti)symmetrization of indices commutes with coordinate transformations, meaning that (anti)symmetrized pieces will retain their form, they'll span invariant subspaces. So the first rule I made was to perform symmetrizations/antisymmetrizations as much as possible.

Next, there are some tensors in theory which are invariant. If we look at general transformations the invariant tensor is Kronecker delta $\delta^\mu_{\hphantom{\mu}\nu}$. There's also Levi-Civita symbol, but that's not a tensor in curved spaces, so I decided to disregard it for now. So as the second rule, we make contractions with these invariant tensors to further reduce the tensor at hand. And with those two rules I planned to proceed.

So if we look at the tensor of 2nd rank $A_{\mu\nu}$, under orthogonal transformations, the irreducible pieces are the trace, traceless symmetric part, and the antisymmetric part:
$$A_{\mu\nu} = \frac{1}{d}Ag_{\mu\nu} + \left(A_{(\mu\nu)} - \frac{1}{d}Ag_{\mu\nu}\right) + A_{[\mu\nu]}$$
where $A = g^{\mu\nu}A_{\mu\nu}$ and $g$ is the metric tensor.
But in the orthogonal case, the metric is preserved under orthogonal transformations, so metric counts as an invariant tensor, but in case of general transformations, this isn't the case. So then maybe this decomposition shouldn't separate the trace and the traceless symmetric part when we're not dealing with orthogonal group? Then again, metric is a special tensor which raises and lowers indices, which means that this composition for a (1,1) tensor would look like:
$$A^{\mu}_{\hphantom{\mu}\nu} = \frac{1}{d}A\delta^\mu_{\hphantom{\mu}\nu} + \left(A^{(\mu}_{\hphantom{\mu}\nu)} - \frac{1}{d}A\delta^\mu_{\hphantom{\mu}\nu}\right) + A^{[\mu}_{\hphantom{\mu}\nu]}$$
and now this conforms to our rules, so it looks like metric should be included as the type of tensor we use in these decompositions.

Am I overthinking this, or these types of constructions should give good results?

 

[fresh_42]

... so you don't really have to pose this question in practice.

This is how far I could follow you.

Either way, I wanted to make a note for myself when I look at general tensors, how to construct irreducible pieces. And so, instead of looking at a particular group, I wanted to look at general coordinate transformations.

And this is where I have no idea what you mean.

A (linear) finite-dimensional representation is a homomorphism. In the case of groups, it is a group homomorphism
$$
\varphi \, : \,G\longrightarrow GL(V)
$$
into the group of regular linear transformations of a finite-dimensional vector space $V.$ It is called irreducible, if $\{0\}$ and $V$ are the only subspaces of $U\subseteq V$ such that $\varphi (g).U\subseteq U$ for all $g\in G.$

You cannot say "forget the group" because that is all that matters. We study the right-hand side, the matrices, in order to understand the left-hand side, the group multiplication. Without the group, you don't have a representation.

My other remark is: where are the tensors you are talking about? You can build the tensor product of different representations, or a representation on tensor spaces, but neither has primarily anything to do with irreducibility.

 

[Antarres]

@fresh_42 Well, the thing is, I used these kinds of decomposition with standard angular momentum(Wigner-Eckart theorem), then in supersymmetry with $SL(2,\mathbb{C})$, and finally in GR with Ricci decomposition of the Riemann curvature tensor.

When I say irreducible tensors, maybe that's not the best terminology, it's the one physicists use, but it's loosely based on representation theory(where I first saw irreducible components as in irreducible representations of a group). So irreducible tensors are said to be invariant(in the sense that they keep the same form under transformations), as well as irreducible(i.e. not reducible, there are no smaller invariant pieces).

The decomposition I gave as an example is the typical example of decomposing a spherical(Cartesian) tensor, which is a tensor that transforms under the group SO(3). But I wanted to make such a decomposition for general coordinate transformation and get for example Ricci decomposition, or a decomposition of any tensor I could get in GR by looking at its symmetry properties.

Then if I would look at group transformations and define tensors with respect to them(Lorentz tensors, SO(3) tensors or any other), I would just refine the definition to include the defining property of the group, such as orthogonality, or whatever property defines the group, and thus obtain the decomposition for this specific group.

So what I meant is that tensors are operators defined under general coordinate transformations, and then these transformations can be representations of some group, but they don't have to be, so I wanted to look at this general case, if that makes sense.

 

[fresh_42]

Maybe this thread should be moved to physics then?

 

[Antarres]

I don't mind that, I just didn't know where to put it and thought it's a linear algebra question since it considers tensors.

 

[fresh_42]

I don't mind that, I just didn't know where to put it and thought it's a linear algebra question since it considers tensors.

The question is linear algebra. The language is physics and makes mathematically no sense unless you specify $G,V,\varphi$ and the terms tensor and coordinates.

 

[Antarres]

Maybe it can be moved to GR? The language used here is used in GR as well. But then again, it's also used in quantum field theory, so not sure.

I guess in physics the terms that are in math different, are kind of conflated, so I'll try to define it more precisely, and if linear algebra is not the appropriate forum for what this results in, feel free to move it elsewhere where it should belong.

So, if we have some finite dimensional vector space $V$, and $V^*$ its dual, the way I would define a tensor usually would be a multilinear functional: $T : V \otimes \dots \otimes V \otimes V^* \otimes \dots \otimes V^* \rightarrow \mathbb{F}$, where $\mathbb{F}$ is the field over which $V$ is defined, take it for simplicity to be real numbers(or complex, doesn't matter I guess), and if the vector space $V$ appears $q$ times in the product, and its dual appears $p$ times, then the tensor is of rank $(p,q)$.

Then for example if we look at GR, we have some smooth manifold as spacetime, this vector space would be tangent space on that manifold. This manifold has charts which define the coordinates and a $(p,q)$ type tensor components transform under coordinate transformations between these charts as:
$$T'^{\mu_1\dots\mu_p}_{\hphantom{\mu_1\dots\mu_p}\nu_1\dots\nu_q} = \left(\frac{\partial x'^{\mu_1}}{x^{\rho_1}}\right)\dots\left(\frac{\partial x'^{\mu_p}}{x^{\rho_p}}\right)\left(\frac{\partial x^{\sigma_1}}{x'^{\nu_1}}\right)\dots\left(\frac{\partial x^{\sigma_q}}{x'^{\nu_q}}\right)T^{\rho_1\dots\rho_p}_{\hphantom{\rho_1\dots\rho_p}\sigma_1\dots\sigma_q}$$

Now this is differential geometry in this case, but in linear algebra this corresponds to a basis transformation, I would say. In physics, the tensor is usually defined as an object that has this type of rule of transformation. When we want to define it corresponding to some group, then these Jacobian matrices are replaced by a representation of the group(and inverse Jacobians by the inverse of this representation matrix). But I wanted to look at irreducible decompositions with respect to transformations of this general form.

Sorry for the confusion, hope it clarified at least a bit about what I mean.

 

[fresh_42]

[Moderator's note: moved from the linear algebra forum.]

We have a language issue here: mathematics versus physics. I moved it here hoping for answers to the OP's questions.

 

[PeterDonis]

Am I overthinking this

Maybe. At any rate, you are making a number of incorrect or not well-defined statements, which indicates to me that you might need to take a step back and start from a more basic level.

we make contractions with these invariant tensors to further reduce the tensor at hand

No, that's not how irreducible pieces of tensors are made.

in the orthogonal case, the metric is preserved under orthogonal transformations

The metric of what? For this statement to have any meaning, you need to specify what metric space you are talking about.

this composition for a (1,1) tensor would look like:
$$A^{\mu}_{\hphantom{\mu}\nu} = \frac{1}{d}A\delta^\mu_{\hphantom{\mu}\nu} + \left(A^{(\mu}_{\hphantom{\mu}\nu)} - \frac{1}{d}A\delta^\mu_{\hphantom{\mu}\nu}\right) + A^{[\mu}_{\hphantom{\mu}\nu]}$$

This is wrong because raising an index with the metric does not in general preserve the symmetry and antisymmetry of the indexes.

irreducible tensors are said to be invariant(in the sense that they keep the same form under transformations), as well as irreducible(i.e. not reducible, there are no smaller invariant pieces).

AFAIK this is wrong; the two terms are not the same, nor are they coextensive.

I wanted to make such a decomposition for general coordinate transformation

A tensor is not the same thing as a transformation. I have no idea what "decomposing" a transformation even means.

If you mean you want to consider tensors of other than 2nd rank, in other vector spaces besides standard Euclidean 3-space, then I don't think there is any general formula for how to decompose a tensor of arbitrary rank in an arbitrary space into its irreducible pieces.

this is differential geometry in this case, but in linear algebra this corresponds to a basis transformation, I would say

Any coordinate chart will define a coordinate basis, but not all bases are coordinate bases. So there are many basis transformations that do not correspond to coordinate transformations.

 

[Antarres]

Maybe I'm looking for something that cannot be done in general, but has to be looked at from case to case, but my idea was to try to figure out some kind of general rule of thumb. Namely, if I have a tensor, and in some literature it is said that it is decomposed into irreducible tensors(with some abuse of notation, since I'm mostly in contact with physics literature, the components of a tensor are usually also called tensor in such articles/books), I want to know how to arrive at such a decomposition. The last such decomposition that I looked at was Ricci decomposition, where Riemann tensor was decomposed into 3 parts, the Weyl part, the scalar component, and the traceless Ricci component(look for example here https://en.wikipedia.org/wiki/Ricci_decomposition).

But this decomposition isn't obviously the only one, for example googling irreducible decompositions would result in a lot of articles/pages about Cartesian tensors, that is, tensors defined with respect to $SO(3)$, group of rotations. For such tensors, the decomposition that I mentioned in my first post is a standard decomposition to scalar, traceless symmetric, and antisymmetric part. It's hard to find one google any decomposition different from this, and the most abstract articles would lead to Wigner-Eckart theorem, and the addition of angular momenta in quantum mechanics(that is, the Clebsch-Gordan decomposition).

I was wondering how to show for example that Ricci decomposition of Riemann tensor is 'irreducible' as it is said, and how to find it for tensors of arbitrary many indices with arbitrary symmetry properties in GR. And how to do the same thing for example with $SU(3)$ in particle theory, or $SL(2,\mathbb{C})$ in supersymmetry etc. I don't mean to find a general formula, but to find some kind of algorithm of how to look at it, if that's possible.

Now to answer your questions. In case what I'm asking for above is maybe impossible, let's just focus on GR as an example, and try to make sense of Ricci decomposition.

The metric of what? For this statement to have any meaning, you need to specify what metric space you are talking about.

Well, since I was talking about tensors, I assume that I have some vector space with a metric which defines a scalar product, and allows us to lower and raise indices, that is, connect the vector space with its dual naturally. Whatever this metric is, in general, if we have a transformation of basis which preserves it, meaning it preserves inner product, this transformation is said to be unitary. If we're dealing with a real vector space, it's called orthogonal. If we have Minkowski space, the tensors would be Lorentz tensors, if it's Euclidean space, it would be Cartesian tensors, but orthogonality is defined the same in either case.

This is wrong because raising an index with the metric does not in general preserve the symmetry and antisymmetry of the indexes.

This, for example, looks strange to me, but maybe just because I'm not considering some pathological cases. But for example take a symmetric tensor:
$$A_{\mu\nu} = A_{\nu\mu} \Rightarrow g^{\mu\rho}A_{\rho\nu} = g^{\mu\rho}A_{\nu\rho} \Rightarrow A^\mu_{\hphantom{\mu}\nu} = A_\nu^{\hphantom{\nu}\mu}$$
How is this incorrect?

Edit: I assume, in hindsight that a simple remark could be made that these two tensors live in different spaces, but there should be a very natural isomorphism between these two spaces, so that the problem sums up to this kind of abuse of notation where we don't pay attention to that. I'm not aware, however, of the cases in GR(or other physics branches), where this distinction is necessary. But maybe there's another remark that I'm not seeing.

AFAIK this is wrong; the two terms are not the same, nor are they coextensive.

What's the correct way to say it then? How are irreducible tensors made?

 

[PeterDonis]

What's the correct way to say it then?

There might not be one, because what you are trying to say might not make sense. I'm not sure because I'm still not sure exactly what you are trying to say. I'm not even sure that you know what you are trying to say. Perhaps it will help to clear up a couple of confusions.

Let's start with "invariant". What does it mean? In the context in which we are working, it means a quantity that remains the same under some set of transformations. For example, in GR, "invariant" is usually used to refer to quantities that remain the same under arbitrary coordinate transformations.

Saying that a tensor is "invariant" is therefore somewhat of an abuse of terminology, because the components of a tensor do not in general stay the same under arbitrary coordinate transformations. Only scalars, i.e., quantities with no free indexes, are invariants if we are going to be careful about terminology. But tensor equations are invariant under arbitrary coordinate transformations: for example, the Einstein Field Equation, $G_{\mu \nu} = 8 \pi T_{\mu \nu}$, is such an equation. It is valid in any coordinate chart; individual components of the tensors involved will change, but the equation remains valid.

Now let's look at "irreducible". What does it mean? It means a representation of a group that cannot be decomposed into "smaller" representations. So to even use the term "irreducible" to refer to tensors, we have to interpret the tensors as representations of a group. Which of course begs the question: what group? And what representation? And, of course, if you are not interested in looking at tensors as representations of a group, then "irreducible" is simply irrelevant to whatever you are trying to do; there is no such thing as an "irreducible tensor" apart from a specific group representation.

 

[Antarres]

Okay, well I guess we agree on both the terms that you mentioned. When saying that something is invariant, it doesn't change under coordinate transformations. It's true that this is not the case for components of a general tensor, but it is true for Kronecker delta, which I assume is the only tensor with that property. Every tensor equation is invariant under coordinate transformations, so that wasn't the sense in which I meant the term 'invariant' but rather the former.

When it comes to "irreducible", I agree that it is a term coming from representation theory. Reading further about Ricci decomposition, it seems that I have wrongly assumed that here no group is at play, but that actually, this decomposition is a decomposition of a tensor with the symmetries of a Riemann tensor, under the action of the orthogonal group. Still the notation used in later part of mathematical explanation on the wiki page that I linked seems a bit obscure for me to understand. I guess in GR in 4 dimensions, this 'orthogonal' group, would be Lorentz group, and the representation would be(tensor product of) fundamental representation, although I don't see that that's written in the article(and the vector space would be the tangent space).

I have wrongly assumed that this particular decomposition has nothing to do with representation theory, but still, without accepting it(the formula) as a fact, I still don't see how I would arrive at it myself.

I guess analogy should be made to how it's done with angular momenta in quantum mechanics, where tensor products of representations are decomposed into a direct sum of irreducible representations, but I'm not sure I see clearly how this is supposed to be done.

What I was giving as an example in the OP was this kind of decomposition under orthogonal group for a 2nd rank tensor that is known to be given by:
$$A_{\mu\nu} = \frac{1}{n}Ag_{\mu\nu} + \left(A_{(\mu\nu)} - \frac{1}{n}Ag_{\mu\nu}\right) + A_{[\mu\nu]}$$
where $g_{\mu\nu}$ are the components of the metric, $A = A^\mu_{\hphantom{\mu}\mu}$ and $n$ is the dimension of the space we're working in.
One of the ways in which we can arrive at this kind of decomposition, is that we notice that the symmetry properties will be invariant under our transformation, so therefore by symmetrizing and antisymmetrizing the tensor, we divide our space of all tensors into two invariant spaces. Further than that, we know that $g_{\mu\nu} = T^{\rho}_{\hphantom{\rho}\mu}T^{\sigma}_{\hphantom{\sigma}\nu}g_{\rho\sigma}$, if $T$ is an orthogonal transformation, because this is the definition of orthogonal transformation. And so $g_{\mu\nu}$ is an invariant tensor, so taking the trace of our tensor and multiplying it by $g_{\mu\nu}$ we obtain the 'trace' irreducible component as it is usually called, and this component acts in a subspace of the symmetric space, so we divide this space into trace + traceless part, and we arrive at the decomposition we have.

Now this is maybe an informal argument, but in such a way a lot of decompositions are verified in physics, when they are provided. My point was that I want to find some kind of rule of the thumb similar to this for the cases when such a decomposition is not provided for verification, but where I'd create it myself. For example if I'm looking at a higher order tensor with some defined symmetry properties. So on the example of Ricci decomposition, I wanted to do the same, but I had doubts about how to do it, and so I went on into this, maybe unclear formulation of my problem in this thread.

 

[PeterDonis]

we know that $g_{\mu\nu} = T^{\rho}_{\hphantom{\rho}\mu}T^{\sigma}_{\hphantom{\sigma}\nu}g_{\rho\sigma}$, if $T$ is an orthogonal transformation, because this is the definition of orthogonal transformation. And so $g_{\mu\nu}$ is an invariant tensor

I'm not following you here. Unless the metric $g_{\mu \nu}$ has the required symmetry property, which a general metric does not, it will not be invariant under orthogonal transformations.

taking the trace of our tensor and multiplying it by $g_{\mu\nu}$ we obtain the 'trace' irreducible component

This is true whether $g_{\mu \nu}$ is invariant under orthogonal transformations or not. The decomposition you wrote down for a (0, 2) tensor (i.e., a tensor with two lower indexes and no upper indexes) is a tensor equation and is valid in any coordinate chart.

I want to find some kind of rule of the thumb similar to this for the cases when such a decomposition is not provided for verification

I'm not aware of any general rules for this, as I've already said. You have to take each case individually.

 

[Antarres]

Isn't a metric supposed to be symmetric? I guess there could be some generalization to make the metric have both symmetric and anti-symmetric parts, but the meaning of the metric in any case should be to measure distance(or I guess in case of metric tensor which acts on vectors, is to measure angles and length), so this property would obviously be lost if we relax the symmetry condition, since then angles would have different measures depending on the order of the vectors, or such. I'm not sure, I've never tried to explore this kind of assumption, but I thought it was quite generic for metric on a manifold to be defined as symmetric, and also for metric in a sense of a bilinear form that induces scalar product/measures distance which is the original definition it comes from.

What I meant was that when a transformation is called orthogonal in the first place, it should mean that it preserves scalar product, so if the equation I provided doesn't hold, then the transformation is not orthogonal, at least not in the particular space with metric that we're looking at.

As for the decomposition I mentioned, yes, this decomposition holds in all coordinate charts of course, but the reason for decomposing any tensor like this, is that these parts relate to different 'actions' so to say, in a sense the space of all tensors is divided to 'irreducible components'. For example when used in fluid flow, these three components give tensors known as expansion, shear and vorticity, which are treated as separate quantities of interest. Similar decomposition when analyzing geodesic congurences leads to expansion, shear and vorticity parameters for geodesics, from which Raychaudhuri equation, and focusing theorem are derived. What I mean is that this decomposition can surely be done for any general 2nd rank tensor in the same form, but I'm not sure the meaning of such decomposition would be preserved(otherwise why give names to those special components anyway?).

I'm not aware of any general rules for this, as I've already said. You have to take each case individually.

Hm, okay, so maybe what I was asking for is not possible. Maybe then such decompositions are stemming from decompositions similar to Clebsch-Gordan where tensor products of general representations are decomposed to irreducible parts, just rewritten in some way in tensor language. It seems to me standard in GR texts to talk about irreducible components of curvature, and in case torsion exists(or non-metricity), extra components appear, etc. but I wondered how they were derived, since they're often just written down as a fact.

 

[PeterDonis]

Isn't a metric supposed to be symmetric?

Yes, the metric in GR is always a symmetric (0, 2) tensor.

the meaning of the metric in any case should be to measure distance

If you plug in coordinate differentials, yes.

(or I guess in case of metric tensor which acts on vectors, is to measure angles and length)

If you plug vectors into the metric you get inner products (or the squared norm if you plug the same vector into both slots).

when a transformation is called orthogonal in the first place, it should mean that it preserves scalar product

Any valid coordinate transformation in GR will preserve all invariants. That includes scalar products.

this decomposition can surely be done for any general 2nd rank tensor in the same form, but I'm not sure the meaning of such decomposition would be preserved

The physical meaning of the decomposition will of course depend on what tensor you are decomposing and what physical meaning that tensor has.

maybe what I was asking for is not possible

That has already been stated several times in this thread.

 

[vanhees71]

Saying that a tensor is "invariant" is therefore somewhat of an abuse of terminology, because the components of a tensor do not in general stay the same under arbitrary coordinate transformations. Only scalars, i.e., quantities with no free indexes, are invariants if we are going to be careful about terminology. But tensor equations are invariant under arbitrary coordinate transformations: for example, the Einstein Field Equation, $G_{\mu \nu} = 8 \pi T_{\mu \nu}$, is such an equation. It is valid in any coordinate chart; individual components of the tensors involved will change, but the equation remains valid.

Saying "a tensor is invariant" is a tautology, because tensors are by definition invariant. In the context of general relativity they are invariant under general diffeomorphisms that transform from one set of (local) coordinates to another.

Tensor components always refer to a basis and the corresponding dual basis and thus transform by specific rules under the change from one set of local coordinates to another such that the tensors themselves stay invariant.

Take as a most simple example a vector $V$ and holonomous coordinate bases $\partial_{\mu}$ and their dual bases $\mathrm{d} q^{\mu}$. Under a diffeomorphism $q^{\mu} \leftrightarrow \bar{q}^{\mu}$ you have
$$\bar{\partial}_{\mu} = \frac{\partial q^{\nu}}{\partial \bar{q}^{\mu}} \partial_{\nu}$$
and
$$\mathrm{d} \bar{q}^{\mu} =\mathrm{d} q^{\nu} \frac{\partial \bar{q}^{\mu}}{\partial q_{\nu}},$$
one says the $\partial_{\mu}$ transform covariantly and the $\mathrm{q}^{\mu}$ contravariantly.

A vector is invariant, and thus its components transform contra-variantly, which is easily shown with the above considerations
$$V=V^{\mu} \partial_{\mu} = V^{\mu} \frac{\partial \bar{q}^{\nu}}{\partial q^{\mu}} \bar{\partial}_{\nu},$$
from which you get
$$\bar{V}^{\nu} = V^{\mu} \frac{\partial \bar{q}^{\nu}}{\partial q^{\mu}},$$
i.e., indeed the $V^{\mu}$ transform contra-variantly.

Now let's look at "irreducible". What does it mean? It means a representation of a group that cannot be decomposed into "smaller" representations. So to even use the term "irreducible" to refer to tensors, we have to interpret the tensors as representations of a group. Which of course begs the question: what group? And what representation? And, of course, if you are not interested in looking at tensors as representations of a group, then "irreducible" is simply irrelevant to whatever you are trying to do; there is no such thing as an "irreducible tensor" apart from a specific group representation.

Now tensors of a specific type build vector spaces, and the linear invertible transformations on any vector space $\mathcal{V}$ form a group, the "general linear group" $\text{GL}(V)$. If you have some group $G$ you can ask for linear representations on a vector space, which by definition is a group homomorphism $\varphi:G \rightarrow \mathrm{GL}(\mathcal{V})$. It's called a homohorphism if $\varphi(g_1 g_2)=\varphi(g_1) \varphi(g_2)$, $\varphi(g^{-1})=[\varphi(g)]^{-1}$. Such a linear representation on $\mathcal{V}$ is called irreducible, if there are no non-trivial subspaces of $\mathcal{V}$, which are left invariant under all $\varphi(g)$ with $g \in G$.

Now consider as the most simple example the rotation group $SO(3)$ in the usual Euclidean vector space $\mathrm{R}^3$. Already its definition is a irreducible group representation on the vector space $\mathbb{R}^3$, i.e., the group consists of the orthogonal $\mathbb{R}^{3 \times 3}$ matrices $\hat{O}$, i.e., matrices for which $\hat{O}^{\text{T}}=\hat{O}^{-1}$ and $\mathrm{det} \hat{O}=1$. This is an irreducible representation.

Of course for any kind of tensor this induces also a representation of SO(3), but these are usually not irreducible. Take, e.g., the 2nd-rank tensors with contravariant components $T^{ab}$. Then under rotations these transform as
$$\bar{T}^{ab}={O^a}_c {O^b}_d T^{cd}.$$
Obviously, however, these transformations leave obviously subspaces of the corresponding 9-dimensional vector space $\mathbb{R}^3 \otimes \mathbb{R}^3$ invariant. E.g., all anti-symmetric tensors, which fulfill $T^{cd}=-T^{dc}$, these build a 3D subspace, which is obviously left invariant under rotations. It's indeed easy to show that $\bar{T}^{ba}=-\bar{T}^{ab}$ if $T^{dc}=-T^{cd}$, i.e., there is a 3D subspace, within which the representation of SO(3) on the 2nd-rank tensors acts without leaving this subspace, i.e., the representation is reducible.

Now you can map the antisymmetric tensor components $T^{ab}$ uniquely to vectors by writing $T^{ab}=\epsilon^{abc} t_c$, where $\epsilon^{abc}$ is the Levi-Civita symbol. Indeed as is well known since the Levi-Civita symbol is invariant under rotations, because
$$\bar{\epsilon}^{def}={O^d}_a {O^e}_b {O^f}_c \epsilon^{abc}=\epsilon^{def} \det \hat{O}=\epsilon^{def},$$
i.e., the rotations act "trivially" on the Levi-Civita symbol. It provides a trivial representation mapping all rotations to the neutral element, which simply doesn't do anything to the Levi-Civita symbol. It's also well-known that in the above mapping of antisymmetric 2nd-rank tensors in $\mathbb{R}^3$ to vectors these vectors transform indeed as they should under rotations, and thus if you restrict the representation of SO(3) to the antisymmetric tensors, that's an irreducible 3D representation of SO(3), i.e., the defining representation again.

Now the same argument holds for the symmetric tensors $T{ab}=T^{ba}$. They transform also again to symmetric tensors under rotations, i.e., these build a 6D subspace, which is left invariant under the SO(3) representations, but that's still not an irreducible representation. That's because you can in addition restrict the symmetric tensors to the trace-less symmetric tensors, which fulfill in addition $T^{ab} \delta_{ab}=0$. Since the trace of a tensor is invariant under rotations, because
$$\bar{T}^{cd} \delta_{cd}={O^{c}}_a {O^d}_b T^{ab} \delta_{cd} =\delta_{ab} T^{ab},$$
where we have used the orthogonality of $\hat{O}$.

One can show that with this restriction you have now a 5D irreducible representation of SO(3) on the vector space of traceless symmetric tensors.

Finally the tensor $\delta^{ab}$ is invariant under rotations (again due to the orthogonality of the SO(3) matrices). It forms thus a 1D (trivial)representation which is of course also irreducible, since a 1D vector space has no non-trivial subspaces.

Now you can decompose any 2nd-Rank tensor into the above constructed irreducible parts. For that you first define

the antisymmetric part
$$A^{ab}=\frac{1}{2} (T^{ab}-T^{ba}),$$
the trace-less symmetric part
$$D^{ab}=\frac{1}{2}(T^{ab}+T^{ba})-\frac{1}{3} \mathrm{Tr} \hat{T} \delta^{ab},$$
which indeed fulfills $D^{ab}=D^{ba}$ and $\mathrm{Tr} \hat{D}=\delta_{ab} D^{ab}=0$ and finally the invariant part
$$I^{ab}=\frac{1}{3} \mathrm{Tr} \hat{T} \delta^{ab}.$$
On each of these parts the rotations are represented by an irreducible representation on the corresponding vector spaces (3D, 5D, and 1d respectively). Thus you have split the 2nd-rank tensors into objects which form vector spaces on which the rotations are represented by irreducible representations,
$$T^{ab}=A^{ab}+D^{ab}+I^{ab}.$$

 

[PeterDonis]

Saying "a tensor is invariant" is a tautology, because tensors are by definition invariant.

That depends on your terminology convention. In some conventions, "invariant" specifically means a scalar invariant, or a quantity composed only of scalar invariants, so there are no components. A quantity with components that transforms appropriately under coordinate transformations, in this convention is called "covariant". That's the convention I was (implicitly) using.

 

[vanhees71]

That's the trouble with the physics literature. Of course, it'll be impossible to change this in any way. Physicists just call components of tensors tensors. That makes it more difficult to understand than necessary. In the modern mathematicians' way the primary objects in linear algebra are vector spaces and linear functions on vector spaces. Bases and components of vectors and tensors come later, and it's almost trivial that the objects, introduced in this way, do not depend on the choice of a basis.

 

[Antarres]

@vanhees71 This is precisely what I was intending to say when giving the decomposition of the 2nd rank tensor under orthogonal group. You made it perfectly clear though, so thank you for that. I wanted, analogously to that, to look to explain the decomposition of the Riemann tensor, that is the Ricci decomposition, as it is called. Since nowhere was any kind of group mentioned when this decomposition was given, at least in any of the literature where I've seen it, I assumed that in this case one should consider, instead of the transformations under some specific group(some Lie group like SO(3)), to consider general coordinate transformations. Now these also form a group(though I never looked at group theory of the group of general diffemorphisms), but I was looking at it using the analogy of the explanation that you made for SO(3). And I was looking to do the same with any type of group, to base the explanation on simple steps that have to be perhaps mildly changed from case to case, but that are in principle the same thing.

Still, after continuing to read about Ricci decomposition, which looked so general to me at first, I realized that my assumption was wrong, and that this decomposition was also related to the orthogonal group, instead of being 'general' as I assumed.

I found it very simple to see that by symmetrizing and antisymmetrizing we divide the space of all tensors into invariant subspaces, like you mentioned. So that meant that even if I have tensors of higher rank, exhaustive implementation of these operations would divide the space into invariant parts. Care should just be taken about intersections of spaces which are made by multiple symmetrizations/antisymmetrizations, i.e. these spaces should be made to be disjoint. And so it looked to me that searching for irreducible parts would necessarily involve all possible steps where we divide into smaller and smaller invariant parts, until we can't divide anymore.

And so my idea was to see that we can do this by using symmetrization/antisymmetrization on indices, and also by identifying what I called 'invariant tensors', by which I meant, tensors whose components don't change under the transformations we're looking at, such as Levi-Civita symbol and Kronecker delta, in case we're looking at 3D rotations in Euclidean space. Using contractions by these tensors and symmetrization/antisymmetrization on indices, I figured that I would obtain all the irreducible parts of some general tensor. And I wanted to check if my logic looks reasonable.

Now I assume that Ricci decomposition was also somehow obtained in that way, and that this way is very general, because it doesn't assume from the start what should be these 'invariant tensors' and those would differ from case to case, but I figured that maybe I'm wrong. It boiled down to my terminology being sloppy, which I was aware of, even when I started posting, but I haven't in literature related to physics found any different terminology, and so I assumed that this kind of terminology is somewhat familiar.

The way you explained it basically makes this precise enough. I'm just wondering if a similar construction can be made for other groups, or higher ranked tensors, etc etc, or it would require a completely different construction. And if it's possible finally, to write down a few cases for myself to use as future reference, since it's more useful to me to know such a principle, than to know by heart properties of all the groups/transformations I work with from case to case, which I'll surely forget if not in contact with the matter during some period.

 

[PeterDonis]

Physicists just call components of tensors tensors.

I'm not sure this is true. I think it's much more common to observe the distinction between tensors, the geometric objects which are basis independent, and their components, which are basis dependent.

 

[samalkhaiat]

1)The Weyl Conformal Tensor:

You really need to know where the Weyl tensor does come from. Weyl obtained his tensor $C_{\rho\sigma\mu\nu}$ from the Riemann tensor $R_{\rho\sigma\mu\nu}$ by subtracting all pieces that are not invariant under arbitrary rescaling of the metric ($g_{\mu\nu}(x) \to \bar{g}_{\mu\nu}(x) =\Omega^{2}(x)g_{\mu\nu}(x)$). This way Weyl obtained $$C_{\rho\sigma\mu\nu} = R_{\rho\sigma\mu\nu} - \hat{S}_{\rho\sigma\mu\nu} + \hat{G}_{\rho\sigma\mu\nu} ,$$ where $$\hat{S}_{\rho\sigma\mu\nu} = \frac{1}{n - 2} \left( g \wedge S \right)_{\rho\sigma\mu\nu}$$
$$\hat{G}_{\rho\sigma\mu\nu} = \frac{R}{2n(n - 1)} \left( g \wedge g \right)_{\rho\sigma\mu\nu},$$ with $$\left(A \wedge B \right)_{\rho\sigma\mu\nu} \equiv A_{\rho [\mu}B_{\nu ]\sigma} - A_{\sigma [\mu}B_{\nu]\rho} ,$$ and the traceless Ricci tensor $$S_{\mu\nu} = R_{\mu\nu} - \frac{1}{n}g_{\mu\nu}R .$$ I cannot reproduce the derivation in here because it is long and tedious, but you can gain so insight about it if you read my post in
https://www.physicsforums.com/threa...nder-conformal-symmetries.998328/post-6442833

Also (for fun), look at

https://www.physicsforums.com/threads/weyl-tensor-on-3-dimensional-manifold.128275/post-1058270

2) Decomposition of the Riemann Tensor (so called “Ricci decomposition”):
This is one of many instances where people associated Weyl work with another person (I have never understood why!) Any way, if you now rearrange the terms in the definition of the Weyl tensor, you get (it is even silly to write it again, so I am ignoring the indices) $$R = C + \hat{S} - \hat{G} .$$ One then can say that the Weyl conformal tensor $C_{\rho\sigma\mu\nu}$, the traceless Ricci tensor $S_{\mu\nu}$ and the Ricci scalar $R$ are the irreducible components of the Riemann tensor $R_{\rho\sigma\mu\nu}$. These are called irreducible components because no new quantities can be obtained from $C_{\rho\sigma\mu\nu}$, $\hat{S}_{\rho\sigma\mu\nu}$ and $\hat{G}_{\rho\sigma\mu\nu}$ by contraction of their indices. The decomposition is called orthogonal because $C \cdot \hat{S} = C \cdot \hat{G} = \hat{S} \cdot \hat{G} = 0$ and $R^{2} = C^{2} + \hat{S}^{2} + \hat{G}^{2}$ where the scalar product is defined by total contractions $A \cdot B = A_{\mu\nu \cdots }B^{\mu\nu \cdots}$. Now the magic happens if we consider the representations of the orthogonal group $SO(4)$ (or the Lorentz group $SO(1,3)$). Since, for these groups, the metric is an invariant tensor, the decomposition can be written (symbolically) as $$R_{\rho\sigma\mu\nu} = C_{\alpha \beta \gamma \tau} \oplus S_{\delta \pi} \oplus R .$$ We know that $C_{\mu\nu\rho\sigma} \in \mathcal{V}^{[5] \oplus [\bar{5}] }$, $S_{\mu\nu} \in \mathcal{V}^{[9]}$ and $R \in \mathcal{V}^{[1]}$ are irreducible representations of the group $SO(4)$. Thus, when referred to the this group, the decomposition represents the following Clebsch-Gordan (vector space) decomposition $$[20] = \left( [5] \oplus [\bar{5}]\right) \oplus [9] \oplus [1].$$ That is to say that the space of tensors having the algebraic symmetries of the Riemann tensor decomposes into irreducible representation spaces of the group $SO(4)$ or $SO(1,3)$. I hope it is clear to you now that the above decomposition has nothing to do with the group of general coordinate transformations. In fact, the Clebsch-Gordon series does not work for the group of general coordinate transformations because this group, unlike semisimple Lie groups, has no invariant tensors. For example, in the Lorentz group, $\eta^{\mu\nu}$ and $\epsilon^{\mu\nu\rho\sigma}$ are invariant Lorentz tensor, and you can use them to define higher rank invariant tensor $$T^{\mu\nu\rho\sigma} = a \eta^{\mu\nu}\eta^{\rho\sigma} + b \eta^{\mu\rho}\eta^{\nu\sigma} + c \eta^{\mu\sigma}\eta^{\nu\rho} + d \epsilon^{\mu\nu\rho\sigma} .$$

 

[vanhees71]

The general covariance of GR is not a physical symmetry but a gauge symmetry, i.e., a redundancy in the description of a physical situation. Indeed, physics doesn't depend on our arbitrary choices of coordinates. That holds true, by the way, for all of physics. You can every theory formulate in a coordinate-independent form. E.g., Newtonian mechanics can be formulated with the Lagrange formalism, which leads to the Euler-Lagrange equations, which are form-invariant under arbitrary choices of "generalized coordinates".

The important physical symmetry of GR is still Lorentz invariance as in SR, only it's made local, which determines the dynamics of the theory.

 

[Antarres]

@samalkhaiat, this makes it a lot clearer to me. I will look at the links you provided for the details, but already the things you mentioned clarified a lot of what I was asking. Thank you!