# Tensors and rotations

All the tensors that I have studied so far have always appeared with some kind of rotation. For example, spherical tensors rotate as spherical harmonics, tensors in the context of special relativity transform via the Lorentz matrices that are just rotations in the the 4 dimensional space-time. My question is the next, do all objects that are called tensors always have to have some kind of rotation associated with it?

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They are two different ways to look at tensors. If you're a linear algebra person, then tensors are just objects (like matrices) that act on a generalised vector space, and there is no notion of rotation (remember that vector spaces can be abstract). But they also form representations of the groups $SO(N)$ and $SU(N)$, which are the rotation groups of real and complex space. So it's indeed natural to link tensors to rotation groups. –  Vibert Jun 1 '13 at 21:32

There is a rather general notion of tensor in physics for which the answer to your question is yes provided you replace the word "rotation" with "group element." In particular, the notion of a tensor that is often used in physics is not restricted to that of multilinear maps on manifolds (which naturally have a particular transformation law under coordinate transormations).

Notice that both the set of rotations, and the set of Lorentz transformations form groups under matrix multiplication. The group of rotations on $3$-dimensional Euclidean space, $\mathbb R^3$, is called $\mathrm{SO}(3)$. The groups of Lorentz transformations on Minkowski space $\mathbb R^{3,1}$ is called $\mathrm{SO}(3,1)$. So when we say that a tensor is an object that transforms in a specific way under rotations, we are specifying that the object transforms in a specific way when acted on by elements of certain groups. For example, a $(k,0)$ tensors under rotations transforms as $$T^{i_i\cdots i_k} \to R^{i_1}_{\phantom{i_1}j_1}R^{i_k}_{\phantom{i_k}j_k}T^{j_1\cdots j_k}$$ We can generalize this to arbitrary groups as follows. Let $G$ be a group, and let $\rho$ be a representation of the group acting on a vector space. This means that $\rho$ assigns a linear transformation $\rho(g)$ on the vector space to each group element $g$. In a given basis for the vector space, we can write the matrix representation $\rho(g)^i_{\phantom ij}$. Then we can define a $(k,0)$ tensor with respect to the representation $\rho$ as an object $T^{i_i\cdots i_k}$ that transforms as $$T^{i_i\cdots i_k} \to \rho(g)^{i_1}_{\phantom{i_1}j_1}\rho(g)^{i_k}_{\phantom{i_k}j_k}T^{j_1\cdots j_k}$$ Notice that tensors under rotations and Lorentz tensors are a special case of this definition when the representations in question are those that map each rotation to itself and each Lorentz transformation to itself (often called the defining representation).

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Any valid tensor equation remains valid under any "nice" change of coordinates, where "nice" means that the both the change of coordinates and its inverse are differentiable and one-to-one. Rotations and Lorentz boosts are just two very specific examples of changes of coordinates. For example, you could do a crazy change of coordinates like $x\rightarrow x'=x+a\sin(b x)$, and as long as $|a|<1$ and $b\ne 0$ (so that the transformation is "nice"), any tensor equation expressed in $(t,x,y,z)$ coordinates will be equally valid in $(t,x',y,z)$.

So no, there is no special link between tensors and rotations.

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What's important when considering the behavior of tensors aren't specifically rotations, but any arbitrary coordinate transformation (a change of basis of your vector space). In general if you have a tensor of rank $(k,l)$, under a change of coordinates, it must satisfy the tensor transformation law: http://en.wikipedia.org/wiki/Tensor#Tensor_fields. Too see this, it is helpful to know the precise definition of tensors as multilinear maps: http://en.wikipedia.org/wiki/Tensor#As_multilinear_maps

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Practically, if you have a coordinate transformation law : $$x \rightarrow y$$ a m-contravariant, n-covariant, tensor $T_{\large i_1...i_n}^{\large j_1...j_m}$ will transform as :
$$T_{\large k_1...k_n}^{\large l_1...l_m} = (\frac{\partial y^{\large l_1}}{\partial x^{ \large j_1}} ...\frac{\partial y^{ \large l_m}}{\partial x^{ \large j_m}}) (\frac{\partial x^{ \large i_1}}{\partial y^{ \large k_1}} ...\frac{\partial x^{ \large i_n}} {\partial y^{ \large k_m}})T_{\large i_1...i_n}^{ \large j_1...j_m}$$