Prove Levi-Civita Symbol is Only 3D Isotropic Tensor

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SUMMARY

The Levi-Civita symbol is established as the only isotropic tensor in three-dimensional space, specifically under the group of spatial rotations represented by SO(3). This conclusion arises from the tensor product of three vectors, which decomposes into a completely antisymmetric part, confirming that the Levi-Civita tensor has one independent component and behaves as a scalar. It is important to note that while it is invariant under SO(3), it is technically classified as a tensor density due to the determinant properties of SO(3) elements. For a detailed proof, refer to "Vectors, Tensors, and the Basic Equations of Fluid Mechanics" by Rutherford Aris, particularly Chapter 2.7.

PREREQUISITES
  • Understanding of isotropic tensors and their properties
  • Familiarity with the special orthogonal group SO(3)
  • Basic knowledge of tensor algebra and tensor products
  • Introduction to group representation theory
NEXT STEPS
  • Study the properties of the Levi-Civita tensor in three dimensions
  • Learn about the decomposition of tensor products in tensor algebra
  • Explore group representation theory, focusing on SO(3) and its applications
  • Read "Vectors, Tensors, and the Basic Equations of Fluid Mechanics" by Rutherford Aris for a foundational understanding
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Students and professionals in fluid mechanics, theoretical physics, and applied mathematics, particularly those interested in tensor analysis and group theory applications in three-dimensional spaces.

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My fluid mechanics textbook says so but gives no proof, I see why it's isotropic but I can't think of why it's the only isotropic tensor in 3D space.
 
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What does isotropic here mean? Invariant under rotations (elements from SO(3) )?

I have the feeling you should be a bit more precise. Usually, one derives invariant tensors under specific group elements from group decomposition. You're working in three dimensional Euclidean space, so you should look at spatial rotations, which are elements of SO(3). For instance, if you denote by V the vector representation of SO(3) and by S the scalar representation, one should have

<br /> V \otimes V \otimes V = S_A + \ldots<br />

This means that the tensor product of three vectors can be decomposed in a completely antisymmetric part (which is the meaning of the subscript A) plus other stuff not important for your question. A completely antisymmetric 3-tensor in 3 dimensions has one independent component (check this!), and hence is "effectively a scalar". This shows that one has an invariant (isotropic!) tensor in three dimensions which is completely antisymmetric: the Levi-Civita 'tensor'.

Note that this is not a tensor for general transformations; one uses the fact that the determinant of an element of SO(3) is +1. Technically, it is a tensor density!
 


thanks, I meant invariant under SO(3). Actually I'm learning some group representation theory now but haven't gone far, so I guess I'll save you post for reading in more details later. And where can I read a detailed proof on this?
 


thanks, I meant invariant under SO(3). Actually I'm learning some group representation theory now but haven't gone far, so I guess I'll save you post for reading in more details later. And where can I read a detailed proof on this?
 


Perhaps it's somewhere in Georgi's text on Lie groups :)
 


I found a elementary proof of this, I don't know if the group method is neater or not, but this one is definitely much more elementary. It's in a book called "Vectors, tensors, and the basic equations of fluid mechanics" by Rutherford Aris, and method can be found in chap 2.7. The basic idea is to first show Levi-Civita is indeed invariant, second by considering a few special rotations to show that if a rank-3 3-D tensor has to be Levi-Civita before it can be an isotropic tensor.
 


Actually I'm a bit curious of the nomenclature of "tensor density".It's obvious why it's called "relative tensor", but what does it have anything to do with density?
 

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