When using covariant tensors in relativity or particle physics, there are some statements that seem like analogues of statements known from linear algebra. For example, if we have a symmetric real-valued rank-2 tensor $A^{\mu\nu}$, there always exists a Lorentz transformation that diagonalizes it, i.e. (using matrix notation) a transformation $\Lambda$ such that $\Lambda A \Lambda^T$ is diagonal. Remembering that a Lorentz transformation $\Lambda$ is defined as a rank-2 tensor with $\Lambda \eta \Lambda^T=\eta$, with $\eta$ the Minkowksi metric, this seems like an analogue of the statement that each symmetric matrix is diagonalizable by conjugation with a rotation matrix (i.e. a matrix $R$ such that $R^T I R=I$).

As another example from particle physics, consider a 4-vector $k^{\mu}$ and choose 3 polarization vectors $\epsilon_i^\mu$ obeying $\epsilon_i^\mu k_\mu=0$, ${\epsilon_i}^\mu{\epsilon_j}_{\mu}=-\delta_{ij}$. Then it can be shown that \begin{equation} \sum^3_{i=1}{\epsilon_i}^\mu{\epsilon_i}^\nu=k^\mu k^\nu/k^2-\eta^{\mu\nu} \end{equation} regardless of the choice of the ${\epsilon_i}^\mu$. If we replace the metric tensor by an identity matrix, this equation looks like it denotes two equivalent ways of writing a projector onto the orthogonal complement of $k^\mu$, up to some minus signs.

So my question is: Is there a more general way to understand these correspondences resp. make them more formal?

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    $\begingroup$ The analogy is right there: what you call "linear algebra" is a vector space with metric $g_{\mu\nu}=\delta_{\mu\nu}$. In SR we use $g_{\mu\nu}=\eta_{\mu\nu}$. $\endgroup$
    – Javier
    Nov 30, 2016 at 19:28
  • $\begingroup$ Why is that puzzling you? Tensor analysis is exactly linear algebra (onto "curved" vector spaces). $\endgroup$
    – gented
    Nov 30, 2016 at 20:00
  • $\begingroup$ My question is rather about more (mathematically precise) context on why such statements are true. For example, how would the lower statement look like for a metric of an arbitrary signature? Can it be generalized further? Is there a trick to reduce the proof of these statements to the case $g_{\mu\nu}=\delta_{\mu\nu}$? What is the name of a "projector onto an orthogonal complement", when using that kind of metric? My understanding of these analogies is pretty handwavy, the proofs of those statements I know of seem "one-off", and there most be some more general theory in which this is true. $\endgroup$
    – Joris
    Nov 30, 2016 at 20:21

1 Answer 1


I don't fully understand your problem, but here is an attempt to answer:

Let $V$ be a finite $n$ dimensional real vector space. Let $e_i$ ($i=1,..,n$) be a basis of $V$. A vector $v\in V$ van be decomposed as $v=v^ie_i$ (summing is implied on repeated indices). A common convention is to put upper indices on components, and lower indices on basis vectors.

Let $V^*$ be the dual space of $V$. Elements of the dual space are linear functionals (real-valued linear maps) on $V$. It is a well known result of linear algebra that if $e_i$ is a basis of $V$, there exists a unique basis of $V^*$ associated to the basis $e_i$, written as $e^i$, such that $e^i(e_j)=\delta^i_j$. A dual vector $\omega\in V^*$ can be uniquely decomposed as $\omega=\omega_i e^i$. For dual vectors we use the opposite convention to vectors, components are labelled with lower indices, and basis vectors are labelled with upper indices.

There reason for this is as so: If $f_j=B^i_je_i$ is a new basis of $V$ ($B^i_j$ are matrix elements of a matrix $B\in\text{GL}(n,\mathbb{R})$), in order for the vector $v\in V$ to stay unchanged, $$ v=v^ie_i=\tilde{v}^jf_j=\tilde{v}^jB^i_je_i, $$ the old components $v^i$ must be related to the new ones as $v^i=\tilde{v}^jB^i_j$. Multiplying by the inverse matrix gets us $$ v^i(B^{-1})^k_i=\tilde{v}^jB^i_j(B^{-1})^k_i=\tilde{v}^j\delta^k_j=\tilde{v}^k, $$ so the new components are related to the old ones by the inverse of the change of basis matrix $B$, hence these components are called contravariant (changing the opposite way). Now, let $f^i$ be the dual basis of $f_i$, then $f^i(f_j)=\delta^i_j=f^i(B^k_je_k)=B^k_jf^i(e_k)$. Since we got $\delta^i_j=B^k_jf^i(e_k)$, so $f^i(e_k)=(B^{-1})^i_k=(B^{-1})^i_j\delta^j_k=(B^{-1})^i_je^j(e_k)$, we have $f^i=(B^{-1})^i_je^j$, so if we want the new bases to respect duality, the dual basis has to change by the inverse of the change of basis matrix, but from that it follows that dual vector components must change by the inverse of the inverse of the change of basis matrix, so by the change of basis matrix.

The conclusion is: If a $f_j=B^i_je_i$ basis transformation is committed, vector components $v^i$ change by the inverse of $B$, dual vector components $\omega_i$ change by $B$, hence the latter are called covariant (changing the same way).

What happens is a dual vector $\omega=\omega_i e^i$ acts on a vector $v=v^je_j$? We compute $\omega(v)=\omega_ie^i(v^je_j)=\omega_iv^je^i(e_j)=\omega_iv^j\delta^i_j=\omega_iv^i$, so this action can be computed by summing the product of components, the basis vectors do not appear. This sum is guaranteed to be independent of the basis chosen, because upper and lower indices change in inverse ways during the change of basis.

Now we discuss tensors. Tensors are unique linearizations of multilinear objects. If $V$ and $W$ are finite dimensional vector spaces over the same field (for example, the reals), then the tensor product of $V$ and $W$ is the pair $(V\otimes W,p)$, where $V\otimes W$ is a real vector space (sometimes also referred to as the tensor product), and $p:V\times W\rightarrow V\otimes W$ is a bilinear map, such that the pair $(V\otimes W,p)$ has the so-called universal factorization property, which says that if $A:V\times W\rightarrow X$ is any bilinear map, then there exists a unique linear map $A^\otimes:V\otimes W\rightarrow X$ for which $A=A^\otimes\circ p$.

Therefore, the tensor product space is such that bilinear maps on the factors are represented uniquely as linear maps on the product space, eg. we can study multilinear problems by studying linear problems on the tensor space. If $v\in V$ and $w\in W$, then $p(v,w)$ is written as $v\otimes w$.

The standard theorem in multilinear algebra states that the tensor product exists and is unque, uniqueness means that if $(V\otimes'W,p')$ is also the tensor product of $V$ and $W$, then there exists a unique canonical isomorphism $\psi:V\otimes W\rightarrow V\otimes'W$ such that $p'=\psi\circ p$, therefore all tensor spaces are equivalent.

The proof of existence is usually given by explicit construction. There are multiple possible ways to construct tensor spaces. Algebraists usually do it via a quotient space I don't want to discuss here. Differential geometers and physicists usually take the tensor space to be a space of multilinear maps as such: the space $B(V^*,W^*)$ consisting of bilinear functionals $T:V^*\times W^*\rightarrow \mathbb{R}$ satisfies the universal property, if $p:V\times W\rightarrow B(V^*,W^*)$ is given as $p(v,w)=v\otimes w\in B(V^*,W^*)$ is the bilinear functional that for any $\omega\in V^*$ and $\eta\in W^*$ gives $v\otimes w(\omega,\eta)=\omega(v)\cdot\eta(w)$.

For finite dimensional spaces $(V^*)^*=V$, so $V^*\otimes W^*$ can be identified with the space of bilinear maps maps $V\times W\rightarrow \mathbb{R}$.

Tensor products can be generalized straightforwardly to multiple factors. In physics and in differential geometry, we usually care about tensor product spaces formed from a vector space $V$ and its dual space $V^*$. The space of $(k,l)$ type tensors over $V$ is the tensor space $$ T^{(k,l)}(V)=V\otimes...\otimes V\otimes V^*...\otimes V^*, $$ where there are $k$ factors of $V$ and $l$ factors of $V^*$. As a multilinear functional it does as follows: If $T\in T^{(k,l)}(V)$ and $\omega_1,...,\omega_k\in V^*$ and $v_1,...,v_l\in V$ then $T$ maps the list $(\omega_1,...,\omega_k,v_1...v_l)$ to the real number $T(\omega_1,...,\omega_k,v_1,...,v_l)$ such that this mapping is separately linear in all variables. If $e_i$ is a basis of $V$ with $e^i$ being the dual basis and the $k$-th $\omega$ is $\omega_k=(\omega_k)_ie^i$ and the $k$-th $v$ is $v_k=(v_k)^ie_i$, then the tensor $T$ acts as $$ T(\omega_1,...,\omega_k,v_1,...,v_l)=(\omega_1)_{i_1}...(\omega_k)_{i_k}(v_1)^{j_1}...(v_l)^{j_l}T^{i_1...i_k}_{j_1...j_l}, $$ where $$T^{i_1...i_k}_{j_1...j_l}=T(e^{i_1}...e^{i_k}e_{j_1}...e_{j_l})$$ are the components of $T$ with respect to the dual basis pairs $e_i$ and $e^i$.

So it is clear, the action of a tensor $T$ can be calculated from the components, the basis vectors do not appear. Moreover, it is also clear because of linearity that during change of basis all upper indices will transform as vector components, while all lower indices will transform as dual vector components.

Now, vector valued tensors! We don't have to fill in all arguments of a tensor of type $(k,l)$. For example, if $A\in V\otimes V^*$, then $A$ maps the pair $(\omega,v)\in V^*\times V$ to a real number written as $A(\omega,v)=A^i_j\omega_i v^j$ in a way that this assignment is linear in both variables separately. What happens if we plug in $v$ but not $\omega$? We get $A(\cdot,v)$ which has a slot left that takes in dual vectors in a linear way. Which means that $A(\cdot,v)$ is a linear functional on $V^*$, so it is an element of $V$. Let us write this as $A(\cdot,v)=A(v)\in V$, in components as $A(v)^i=A^i_jv^j$.

We can view elements of $V\otimes V^*$ as old-fashioned linear operators on $V$! The usual stuff about linear operators all apply, including eigenvalue-problems etc.

Likewise, the good old bilinear functionals/forms from linear algebra are just elements of $V^*\otimes V^*$!

In tensor calculus we usually allow working on a manifold as opposed to a vector space (even if this manifold in question ends up being a vector space, such as euclidean 3-space or Minkowski-space). On a manifold, we can only have vectors and tensors at fixed points, there are no "free vectors". So the vector spaces of interest are the tangent spaces $T_pM$ and cotangent spaces $T^*_pM$ defined for each $p\in M$ (points of the manifold).

These are $n$-dimensional real vector spaces (if the manifold is also $n$ dimensional) that serve as local linearizations of the manifold. If $(x^1,...,x^n)$ is a coordinate system that is defined around $p$, then the differential operators $\partial/\partial x^i$ provide a natural basis for $T_pM$, while the differentials $dx^i$ provide the dual basis in $T^*_pM$.

If we change from $(x^1...x^n)$ to $(y^1,...,y^n)$, this induces a change of basis $$\frac{\partial}{\partial y^j}=\frac{\partial x^i}{\partial y^j}\frac{\partial}{\partial x^i},$$ so $B^i_j=\partial x^i/\partial y^j$. The inverse matrix is simply $\partial y^i/\partial x^j$.

A $(k,l)$-type tensor at $p\in M$ is just an element of $T^{(k,l)}(T_pM)$. All pointwise tensorial operations (eg. not differentiation of fields) can be performed solely by the components, so if $T\in T^{(k,l)}(T_pM)$, then in the coordinate system $(x^1,...,x^n)$ it can be represented as an array of numerical components $T^{i_1...i_k}_{j_1...j_l}$ that during a coordinate change $x\mapsto y$ gets a factor of $\partial x^i/\partial y^j$ for each lower index and a factor of $\partial y^i/\partial x^j$ for each upper index.

Conclusion: Tensor calculus is doing linear algebra separately in each point of a manifold in a way that you use component representations of linear algebraic objects instead of the abstract objects themselves.

  • $\begingroup$ I am sorry, thanks for writing this up. But it was not really the answer I was looking for. Maybe the question just doesn't have a meaningful answer. $\endgroup$
    – Joris
    Dec 14, 2016 at 15:07

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