I Pseudotensors in different dimensions

illuminates

In this topic https://physics.stackexchange.com/questions/129417/what-is-pseudo-tensor one answer was the next:

The action of parity on a tensor or pseudotensor depends on the number of indices it has (i.e. its tensor rank):

- Tensors of odd rank (e.g. vectors) reverse sign under parity.
- Tensors of even rank (e.g. scalars, linear transformations, bivectors, metrics) retain their sign under parity.
- Pseudotensors of odd rank (e.g. pseudovectors) retain their sign under parity.
- Pseudotensors of even rank (e.g. pseudoscalars) reverse sign under parity.

But I don't understand one thing. Is that statement only for Euclidean three dimensions? I attempted understand it myself. And it is my thoughts. Pseudotensor is determined as:
$$\hat{P}^{i_1\ldots i_q}_{\,j_1\ldots j_p} = (-1)^A A^{i_1} {}_{k_1}\cdots A^{i_q} {}_{k_q} B^{l_1} {}_{j_1}\cdots B^{l_p} {}_{j_p} P^{k_1\ldots k_q}_{l_1\ldots l_p}$$
where $(-1)^A = \mathrm{sign}(\det(A^{i_q} {}_{k_q})) = \pm{1}$
Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{pmatrix}
And $(-1)^A=-1$
Let's consider a pseudovector in Euclidean three dimensions. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 &0 \\
0 & 0 & 0 & -1
\end{pmatrix}
And $(-1)^A=1$
Let's consider a pseudovector in Minkovski space. Then $\det(A^{i_q} {}_{k_q})$ is
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 &0 \\
0 & 0 & 0 & -1
\end{pmatrix}
And $(-1)^A=-1$
am I right?

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Orodruin

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The action of parity on a tensor or pseudotensor depends on the number of indices it has (i.e. its tensor rank):

- Tensors of odd rank (e.g. vectors) reverse sign under parity.
- Tensors of even rank (e.g. scalars, linear transformations, bivectors, metrics) retain their sign under parity.
- Pseudotensors of odd rank (e.g. pseudovectors) retain their sign under parity.
- Pseudotensors of even rank (e.g. pseudoscalars) reverse sign under parity.

But I don't understand one thing. Is that statement only for Euclidean three dimensions? I attempted understand it myself. And it is my thoughts.
You have to be careful with what you call "parity" here. Typically there are many different transformations that take left-handed systems to right-handed ones (for example, if you look at the group O(n) as transformations of Euclidean n-dimensional space, the entire set $R_1 SO(n)$, where $R_1$ is the reflection operator for the first dimension, does this). Now, in odd dimensions, the reflection of every direction is not a proper rotation. However, in even dimensions, the corresponding transformation is actually in SO(n) as the composition of two reflections is a proper rotation. The above is only true if you interpret "parity" only as the transformation that flips all directions, and that is only an improper rotation for odd-dimensional spaces. Therefore, in four dimensions,
\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 &0 \\
0 & 0 & 0 & -1
\end{pmatrix}
is actually a proper rotation. Parity is not related to this, it is related to what happens under improper rotations. The actual difference of how what is a tensor and what is a pseudo-tensor is given by the fact that the components of a tensor transform according to
$$\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}} T^{i'_1 \ldots}_{j'_1 \ldots} = \dd{x'^{i'_1}}{x^{i_1}} \ldots \dd{x^{j_1}}{x'^{j'_1}} \ldots T^{i_1 \ldots}_{j_1\ldots},$$
whereas the components of a pseudo-tensor transform according to
$$T^{i'_1 \ldots}_{j'_1 \ldots} = \mathscr D \dd{x'^{i'_1}}{x^{i_1}} \ldots \dd{x^{j_1}}{x'^{j'_1}} \ldots T^{i_1 \ldots}_{j_1\ldots},$$
where $\mathscr D = \operatorname{sgn}(\det(\partial x'/\partial x))$.

illuminates

in odd dimensions, the reflection of every direction is not a proper rotation
May you explain why?

Orodruin

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Because the determinant of the transformation matrix is one, not minus one. In 2n dimensions you can construct it by making n consecutive two-dimensional rotations by 180 degrees.

illuminates

Because the determinant of the transformation matrix is one, not minus one. In 2n dimensions you can construct it by making n consecutive two-dimensional rotations by 180 degrees.
For example vector of magnetic field is pseudo-vector and it is determined in three-dimension, isn't it?

Sorry, I don't quite understand your first message.
For example in this book http://farside.ph.utexas.edu/teaching/em/lectures/node120.html , author works in Minkovski space and uses parity inversion (x,y,z)->(-x,-y,-z) and comes to the conclusion that $\epsilon^{\mu\nu\rho\sigma} -> - \epsilon^{\mu\nu\rho\sigma}$

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Orodruin

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The "magnetic field" is relevant as an entity only if you restrict the Lorentz transformations to rotations and parity transformations, i.e., you forget about boosts. It is not a 4-vector, but a part of the rank 2 field tensor together with the electric field. If you include boosts the magnetic field mixes with the electric field. It is true that the permutation symbol is a pseudo-tensor, or more precisely, under the improper Lorentz transformations it transforms with a relative sign compared to how a tensor would transform. The improper Lorentz transformations are all Lorentz transformations with a determinant -1, not just those flipping the sign of all spatial coordinates (this is also not a Lorentz invariant statement). The appropriate thing to consider are the different connected components of the Lorentz group.

parity inversion (x,y,z)->(-x,-y,-z) and comes to the conclusion that ϵμνρσ−>−ϵμνρσ
Again, it must be stressed vigorously that this is not the definition of what it means to be a pseudo-tensor.

illuminates

Again, it must be stressed vigorously that this is not the definition of what it means to be a pseudo-tensor.
Are you state that tensor Levi-Civita isn't pseudo-tensor?

May you advise literature where is a discussion about the difference between even-dimensional and odd-dimensional pseudo-tensor and how it links with parity? Because I did't get that.

Orodruin

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To be more specific, if you had a (proper) 4-vector $(V^\mu) = (V^0,V^1,V^2,V^3)$ under Lorentz transformation and did that transformation, it would transform only with the transformation matrix and become $(V^{\mu'}) = (V^0,-V^1,-V^2,-V^3)$. However, if you had a pseudo-4-vector $(W^\mu) = (W^0,W^1,W^2,W^3)$, it would transform with an additional minus sign since the transformation is an improper Lorentz transformation. It would therefore transform to $(W^{\mu'}) = (-W^0,W^1,W^2,W^3)$. Note that if you restrict the Lorentz transformations exclude boosts, the time component here transforms exactly like a pseudo-scalar and the spatial components as the components of a pseudo-vector in three dimensions.

Are you state that tensor Levi-Civita isn't pseudo-tensor?
No. I am stating that what you quote is not the proper definition of what constitutes a pseudo-tensor. It is based on a misunderstanding of what it means to be a pseudo-tensor. It is not a pseudo-tensor because it gets a minus sign under that particular transformation. It is a pseudo-tensor because it gets a relative minus sign under any improper Lorentz transformation relative to how it would transform if it were a proper tensor.

For the Levi-Civita symbol it boils down to the same thing in the end since its components do not change under proper Lorentz transformations. This is not true for the general tensor or pseudo-tensor.

May you advise literature where is a discussion about the difference between even-dimensional and odd-dimensional pseudo-tensor and how it links with parity? Because I did't get that.
It is discussed in Chapter 4.6.2 of my book.

illuminates

To be more specific, if you had a (proper) 4-vector (Vμ)=(V0,V1,V2,V3)(V^\mu) = (V^0,V^1,V^2,V^3) under Lorentz transformation and did that transformation, it would transform only with the transformation matrix and become (Vμ′)=(V0,−V1,−V2,−V3)(V^{\mu'}) = (V^0,-V^1,-V^2,-V^3). However, if you had a pseudo-4-vector (Wμ)=(W0,W1,W2,W3)(W^\mu) = (W^0,W^1,W^2,W^3), it would transform with an additional minus sign since the transformation is an improper Lorentz transformation. It would therefore transform to (Wμ′)=(−W0,W1,W2,W3)(W^{\mu'}) = (-W^0,W^1,W^2,W^3).
May you explain why happens such things. I thought that $V^μ \rightarrow V^μ$ if $V^μ$ is a 4-vector and $W^μ \rightarrow -W^μ$ if $W^μ$ pseudo-4-vector

I caught your statement about pseudo-tensors have to change their sign in odd dimension, but you nothing said about how it is connected with a rank of a tensor.

Will it be more correct statement if one says:
- Pseudotensors of odd rank in Euclidian space retain their sign under improper rotation.
- Pseudotensors of even rank in Euclidian space reverse sign under improper rotation.

?

Orodruin

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Again, the nomenclature with pseudo-tensors "reversing sign" under parity transformations is strongly misleading. You would do best to stop thinking about it that way as it will only lead you astray. It is correct only in odd dimensions and only when you restrict the "parity transformation" to mean $\vec x \to -\vec x$.

The way pseudo-tensors are defined is that they transform with a relative minus sign to how tensors transform under transformations that change spatial orientation, i.e., transformations that change right-handed systems to left-handed ones and vice versa. In even spatial dimensions, the transformation $\vec x \to -\vec x$ is not a parity transformation as it is a proper rotation. Tensors and pseudo-tensors will therefore transform in exactly the same way under these transformation. As an example, a proper rotation in two dimensions can be described using the transformation matrix
$$A(\theta) = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)\end{pmatrix}.$$
As you will realise, this transformation takes $\vec x \to - \vec x$ if $\theta = \pi$ and it is therefore not a parity transformation.

illuminates

To be more specific, if you had a (proper) 4-vector (Vμ)=(V0,V1,V2,V3)(V^\mu) = (V^0,V^1,V^2,V^3) under Lorentz transformation and did that transformation, it would transform only with the transformation matrix and become (Vμ′)=(V0,−V1,−V2,−V3)(V^{\mu'}) = (V^0,-V^1,-V^2,-V^3).

May you explain way Paul Colby in theme https://www.physicsforums.com/threads/vectors-in-minkowski-space-and-parity.937349/ said me that
Time reversal changes the sign of the T bits.

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illuminates

May you explain why you are making several threads on the same subject? This is against forum rules and just serves to fragment the discussion. We do not appreciate it. Also I do not appreciate your misrepresentation in that thread of what I have been saying in this thread.
When I created themes I thought that it will be different topics. I wish to emerge themes but I cannot do it. I left the message in that thread because I thought that Paul Colby will be interested in this thread too but I apparently understand you not right... I'm sorry! I thought that you said about TP reversal. I changed the message in that thread for more appropriate.

Paul Colby

Gold Member
May you explain way Paul Colby in theme
I don't think Orodruin need ever be accountable for what I say. "The T bits" I was referring to the time component of 4-vectors. In the language I'm familiar with, $V^\mu \rightarrow -V^\mu$ is not a parity transformation. What I said applies to a 4 dimensional space-time. In 11 dimensions you're on your own.

Mentor

PeterDonis

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I thought that $V^μ \rightarrow V^μ$ if $V^μ$ is a 4-vector and $W^μ \rightarrow -W^μ$ if #W^μ## pseudo-4-vector
You thought incorrectly. You appear to have a number of conceptual confusions, which are not helping your understanding. These confusions have already been mentioned in this thread, but I'll try to summarize them here.

First: a Euclidean space is not the same as a Minkowskian spacetime. In a Euclidean space, a parity transformation reverses the sign of all the components of a vector (at least, that's the most common meaning for that term that I'm aware of). In a Minkowskian spacetime, a parity transformation only reverses the sign of the space components of a vector; it does not change the sign of the time component. There is a separate transformation called "time reversal" that reverses the sign of the time component but not the space components.

Second: as @Orodruin has pointed out, a parity transformation is not the same as an improper rotation, and the correct definition of a pseudovector is that its components change sign under an improper rotation, not under a parity transformation. In Euclidean 3-space, it happens to be the case that a parity transformation is also an improper rotation; but you cannot generalize that to all cases.

At this point I am leaving the thread closed since the questions you asked in this thread, and in the previous one you started (which I have also closed), have been addressed. I strongly suggest that you take some time to learn more details about these topics.

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