# A fundamental question about tensors and vectors [closed]

Studying relativity, I am deeply confused with the fundamental concept of vectors and tensors. Are they some specific "realities" that "exist" independently of coordinates? If so, given a vector $\textbf{V}$, its covariant components and contravariant components in a certain coordinate system are just the expression of the "reality" in the coordinate system? Also, the components of $\textbf{V}$ in some other coordinates are also just another expressions of the "reality"?

Does the above contents hold for some tensor $\textbf{T}$ of rank $n$ also? For example, is it a true statement that the components of a rank 3 tensor $\textbf{T}$ can be expressed as $T^i_{jk}$, $T^{ij}_k$, $T_{ijk}$...etc in a given coordinate system? Then the word such as "(1,2)-tensor" just apply for the expression $T^i_{jk}$ of $\textbf{T}$?

Besides, given a tensor expression $T^i_{jk}$, is it always true that when lowering the upper index i, $T^i_{jk}$ becomes $T_{ijk}$? Are there cases in which it becomes $T_{jik}$ or such? In other words, when upper indices are lowered, then are the upper indices always placed in front of the existing lower indices?

Here an arbitrary $n$-dimensional space (thus can be applied to the spacetime) is assumed. Thus the index $i$ and $j$ can range from 1 to $n$. $\textbf{L}$ is a transformation to another coordinate system for contravariant indices and $\overline{\textbf{L}}$ is a transformation to another coordinate system for covariant indices. Here what do $g'_{ij}$ stands for? Shouldn't it be the same metric as $g_{kl}$ expressed in another coordinate? But I have a vague memory that $g'_{ij}$ is the metric components for the metric of the coordinate system denoted by '. In the same context, what do $\textbf{g}$ and $\textbf{g}'$ stand for? Are they metric tensor "realities" for the same coordinate system denoted by no prime? Or are they metric tensor realities for the different coordinate systems? I am totally confused.....

• Check that this question is like asking "Do numbers exist?". You won't see "a number" running down the street. All you can see are "representations" of them. The number "10" is "A" in hexadecimal basis... But the idea of "ten units" remains inside... so they exists... but they are abstract entities. – FGSUZ Mar 18 '18 at 12:00
• Please do not post images of texts you want to quote, but type it out instead so it is readable by all users and so it can be indexed by search engines. For formulae, use MathJax instead. – Emilio Pisanty Mar 18 '18 at 13:21
• What is the source of the text that you posted as an image? What is the context? – Ben Crowell Mar 19 '18 at 1:01

The rest of this is discussed above,

You should be careful about the order of the tensor indices when it can be important. So, you shouldn't write $T^{i}_{jk}$ but rather $T^{i}{}_{jk}$, unless you really intend to write $T_{j}{}^{i}{}_{k}$, precisely for the ambiguity you discuss in your last paragraph

• So, what you mean is that expressions like $T^i_{jk}$ are ambiguous? I should write like $T^i\;_{jk}$ or $T_j \;^i \;_k$ in order to be certain? – Keith Mar 15 '18 at 19:04
• But isn't $T^{i}_{jk}$ usually regarded as $T^i \;_{jk}$, if there are no other contexts to specify it? – Keith Mar 15 '18 at 19:07
• @Keith : I would read it like you did in your second column, but in everything I've written, I've made an effort to make the index order unambiguous. – Jerry Schirmer Mar 15 '18 at 19:30
• @Keith : please just submit another question. – Jerry Schirmer Mar 18 '18 at 14:10

Tensors are best thought of as linear maps from an ordered set of vectors and covectors to the reals. In physics, some quantities are born "naturally" with a certain index structure, e.g. a tangent vector to a curve $v^a$ or the metric $g_{ab}$ or electromagnetic field tensor $F_{ab}$ or Riemann curvature tensor $R_{abc}{}^d$ [where I am using the abstract-index notation.. so ${}_a$ is not an index to be summed over, but is a label corresponding to a slot in the ordered set]. (An example from classical mechanics... a conservative force $f_a=-\nabla_a U$ is a covector-field.)

Components of tensors can be thought of as the results of measurements of the tensor by various instruments ("measuring apparatus"). Transformations of tensor components can then be thought of as mapping one set of measurements of the same tensor from one apparatus to another.

When there is an invertible metric, "indices of tensors may be raised and lowered" with the metric. For example, the tensor $v^a g_{ab}$ is often written as $v_b$ and the tensor $R_{abc}{}^{d}g^{ce}$ is written as $R_{ab}{}^{ed}$ (both of which hides the use of the metric). One could argue that $R_{abc}{}^{d}$ [defined by a derivative-operator via $(\nabla_a\nabla_b-\nabla_b\nabla_a)\omega_c=R_{abc}{}^d\omega_d$, for all $\omega_c$] is more fundamental (or more primitive) than $R_{ab}{}^{ed}$ [which needs both a derivative-operator and an invertible metric]. (When there is enough symmetry around, one can exploit it... but sometimes that clouds what the fundamental structures are.)

• The metric tensors themselves are the most confusing ones that are hardest to understand. Could you please answer for my additional question above? – Keith Mar 18 '18 at 9:05

Let's start with a simpler geometry in which the height of indices doesn't matter, namely Euclidean geometry because there the metric is just the identity matrix. You'll be familiar with the fact that a location in space doesn't change just because I rotate Cartesian axes or even switch to some other coordinate system such as a polar one, but the components themselves change because $V_i=V\cdot e_i$ is a dot product with a basis element. Another important fact about vectors is their dot product is invariant under a change in coordinates, so for example if $V,\,W$ are vectors then changing coordinates will affect their components but nor $V\cdot W=V_i\delta^{ij}V_j$ with implicit summation over repeated indices.

Rank-$n$ tensors generalise this to "vectors" with $n$ indices. In fact the word vectors doesn't need square quotes, because the rank-$n$ tensors form a vector space (i.e. they're a set satisfying the vector space axioms). The "dot product" generalises to contracting away some or all repeated indices, possibly adding in metric tensors (such as the above Kronecker delta) to get repeated indices in the first place, leaving a tensor of some rank or another. If the rank is $0$, you have a scalar and hence an invariant as in the paragraph above. A tensor of non-zero rank satisfies a suitable transformation rule for what happens to its components when you change coordinates, and that also generalises from the case of "familiar" vectors i.e. rank-1 tensors.

Explicitly we have $V_i=\Lambda_i^{j'} V_{j'}$ for switching from the $j'$ coordinate to the $i$ coordinate, where e.g. special relativity uses a Lorentz transformation $\Lambda$. Similarly, $V_i=g_{ij}V^j$ can also be thought of as a notational switch; for example, the Riemann tensor is rank 4, but it's not "really" $(1,\,3)$ instead of $(0,\,4)$, $(2,\,2)$ etc.

So in short, tensors are "objective", but their components are coordinate-choice consequences. However, physical laws are written in a form that doesn't vary with coordinates, e.g. (working with $c=1$ for cleaner notation) $m^2 =p^\mu p_\mu=m^2 p^\mu g_{\mu\nu} p^\nu$.

• How about the metric tensors themselves? It is the most confusing part for me. If tensors are "reality"(in my expression) or "objective"(in your expression), does the same fact holds for metric tensors? However, a metric tensor is defined and based on a given "coordinate system". So, a metric tensor is some counterexample showing that tensors themselves or their own "reality" can be dependent on a coordinate system? Could you explain for me? – Keith Mar 18 '18 at 8:46
• Related to this question, I added some more contents to my original one. Could you please answer me? – Keith Mar 18 '18 at 9:02
• @Keith All tensors of a given number of upper and lower induces transform the same way under coordinate transformations, including the metric tensor. In each case the transformation describes the same object with different numbers. Fixing one coordinate system, $g_{\mu\nu}dx^\mu dx^\nu$ is invariant; in fact you can use this to derive how the metric transforms. – J.G. Mar 18 '18 at 9:18
• You mean all tensor "components" with a given number of upper and lower indices. So, is what you are saying is that in the picture I uploaded above, $g'_{ij}$ and $g_{ij}$ denote the same metric tensor? – Keith Mar 18 '18 at 9:33
• And how about the fact that a metric tensor is defined and based on a given coordinate system? Does it counter the argument that tensors are "objective"? – Keith Mar 18 '18 at 9:34

It is assumed that the form of the physical laws should be the same disregarding the coordinates system in which they are stated. The reasoning behind is that the coordinates are just artifices of the human mind. It happens that an equation written in the tensorial formalism is invariant if you change coordinates. For instance the norm of a vector is an invariant, i.e. it is a reality independent of the coordinates in which you measure it. Instead its components, whether contravariant or covariant, depend on the coordinates chosen. As for the raising or lowering of indices, you use the metric tensor in that coordinates system. The indices should however be written in sequence.