Talk:Invariants of tensors

message: Following the comment below, I have to concur that there are serious issues with this page that I will now attempt to address. There is considerable redundancy in information and potentially incorrect statements. There are also factually incorrect statements. For example, the correspondence to the coefficients of the characteristic polynomial only holds true for the principal invariants because there also exists mixed invariants of rank 2 tensors. I do not have the time to change the article completely to a form I think is appropriate, but I will attempt to polish things up.  — Preceding unsigned comment added by MJASmith (talk • contribs) 11:14, 24 January 2019 (UTC)
thread_title:  Seriously deficient and non-reliable information 

message: As far as I can tell, no one has fixed the issues brought up as far back as 2005. In that same vein, I would like to note that the definition of the invariants given in "Properties" as +/- the determinant seems incorrect to the casual observer (me). 
Jdc2179 (talk) 12:35, 7 June 2013 (UTC)
thread_title:  Still Seriously Deficient 

message: This article deals only with rank-2-tensors, i.e. square matrices. 
However, this restriction is not stated anywhere.
One can define invariants for tensors of rank n>2, such as Wigner invariants, but the theory gets considerably more involved.  — Preceding unsigned comment added by 193.174.246.169 (talk) 12:45, 31 January 2013 (UTC)
thread_title: Rank-2-tensors only

message: The current (November 12, 2005) version is seriously misleading:
*article fails to explain clearly difference between tensors as multilinear operators (concept in multilinear algebra) and tensor fields (concepts of tensor calculus); both individual tensors and tensor fields are said to possess invariants, but these concepts exist at distinct levels of structure,
*the terms I_A, \, II_A (first invariant, second invariant, etc.), for the coefficients a_1, \, a_2 in the characteristic polynomial
: p_A (\lambda) = \lambda^n - a_1 \, \lambda^{n-1} + a_2 \, \lambda^{n-2} \dots
of a linear operator (second rank tensor) A is not standard in applied math (or physics); although it may be standard for all I know in some disciplines math usage should trump others, as these are general mathematical concepts useful in many fields,
*the coefficients of the characteristic polynomial of a tensor (concept in tensor algebra) are indeed invariants, but they are by no means the only invariants or even the most important ones (see any math book on invariant theory for example, noting that bilinear forms are associated with linear operators and hence a special case of tensors),
*the dimension counting argument given in the present version fails for more transformation groups than SO(3), but invariant theory in math/physics certainly deals with many many more groups than this one!
I added a link to an article I wrote which also discusses coefficients of the characteristic polynomial, hopefully successfully informing a wide audience of things which everyone who uses these invariants should probably know.---CH  (talk) 23:27, 12 November 2005 (UTC)
thread_title: Serious deficiencies

message: Can someone please fix the latex math renderings to render every equation. Also, should the last section read "...thus only 3 degrees of freedom..." rather than "...thus only degrees of freedom..."?
:Just fixed it. Tomeasy (talk) 10:06, 25 April 2008 (UTC)
thread_title: Engineering application

message: hi guys, seeing some parallels wrt the invariants of a diagonal matrix a la sylvester's law of inertia, and i am thinking maybe it'd be useful to expand upon the properties section and maek the relationship between the two clear?
it would also help because it introduces an explicit link between tensors, invariants, and dr sylvester's law of inertia.
tyvm
96.52.168.137 (talk) 22:26, 16 November 2015 (UTC)
thread_title:  tying properties section in w/ dr sylvester's law of inertia 

message: I guess I would be confusing, since I_i is used for invariants, but would not \boldsymbol{I} or \boldsymbol{1} be sufficiently clear? I've just never seen \boldsymbol{E} used for the identity tensor, and I feel like more common notation would make it more readable.
Justin Kunimune (talk) 17:23, 9 July 2018 (UTC)
thread_title:  Is there a reason '''''E''''' is used for the identity tensor? 

message: Here introduced J_1,\dots,J_3 defined as coefficients of the characteristic polynomial of the deviator \mathbf{A}_\mathrm{dev}:=\mathbf{A} - (\mathrm{tr}(\mathbf{A})/3)\mathbf{I} has completely wrong result, see
G. Holzapfel: "Nonlinear Solid Mechanics: A Continuum Approach for Engineering" (1.180)-(1.181), or Comsol Structural mechanics Module User Guide
In fact for \mathbf{X}\in\mathbb{R}^3, the characteristic polynomial defined as 
:
p_\mathbf{X}(\lambda)=\mathrm{det}\left(\mathbf{X}-\lambda\mathbf{I}\right)
=-\lambda^3+I_1(\mathbf{X})\lambda^2 - I_2(\mathbf{X})\lambda + I_3(\mathbf{X})
we have that
:
\begin{align}
  J_1 &:= I_1(\mathbf{A}_\mathrm{dev}) = 0 \\
  J_2 &:= I_2(\mathbf{A}_\mathrm{dev}) = I_2(\mathbf{A}) - \frac{1}{3}I_1^2(\mathbf{A})\\
  J_3 &:= I_3(\mathbf{A}_\mathrm{dev}) = I_3(\mathbf{A}) - \frac{1}{3}I_1(\mathbf{A})I_2(\mathbf{A}) + \frac{2}{27}I_1^3(\mathbf{A})
\end{align}
All here stated formula were checked using symbolic computation tool. Oldrich Vlach (talk) 08:51, 10 November 2023 (UTC)
thread_title:  Factual error in "Main invariants" 

message: A "rank two tensor" is a just a matrix and this article is almost entirely about matrices invariant under orthogonal rotations aka euclidean space aka the eigenvalue-eigenvector problem. The only mention of actual tensors is that a single sentence that "you can do that too", which is pretty bogus, given that there are entire books written on homogeneous spaces and Casimir invariants and whatnot. This article is wildly misnamed and misleading. It deserves a prod or an AfD or merge-to something else. 67.198.37.16 (talk) 00:42, 19 February 2024 (UTC)
thread_title:  What the heck?