What Is a Tensor?

Overview¶

The word tensor admits at least two standard definitions. One is algebraic: a tensor is an element of a tensor‑product of a vector space with itself and its dual. The other is geometric: a tensor is a field on a manifold that transforms in a prescribed way under changes of coordinates. Both definitions are used in the literature, and both appear in our own notes — Differential Forms uses the algebraic definition, while the differential-geometry chapter of the Theoretical Physics Reference uses the geometric one. This document explains the relationship between the two, and then uses that relationship to clarify a number of related notions: vector, matrix, array, linear transformation, and pseudotensor.

The algebraic definition¶

Following Differential Forms, a tensor is built from a finite‑dimensional real vector space $V$ with dual $V^*$ by taking tensor products. The space of $(r,s)$‑tensors is

equivalently the space of multilinear maps $\underbrace{V^*\times\cdots\times V^*}_{r}\times \underbrace{V\times\cdots\times V}_{s}\to\R$.

Choosing a basis $\{\vec e_i\}$ of $V$ and the dual basis $\{\tilde e^i\}$ of $V^*$ (related by $\tilde e^i \cdot \vec e_j = \delta^i{}_j$), a rank‑2 tensor can be written

a rank‑3 tensor

and a tensor of mixed type for instance

Vectors $\mathbf{u} = u^i \vec e_i$ and forms $\mathbf{p} = p_i \tilde e^i$ are the rank‑1 cases. In index notation the basis is dropped and the tensor is represented by its components alone — $u^i$, $p_i$, $A^{ij}$, $A^{ijk}$, $A^i{}_j$, $p_i u^i$, and so on. Upper and lower indices are just the contravariant and covariant components of the same tensor.

This definition is purely algebraic: it needs only a vector space and its dual, no notion of a manifold, no coordinates, no transformation law.

The geometric (manifold) definition¶

The differential-geometry chapter of the Theoretical Physics Reference instead defines tensors by their transformation rule under a change of coordinates on a manifold. The construction proceeds inductively.

Start with a differentiable manifold $U$ and a coordinate change $x'=x'(x)$, $x=x(x')$.

• A scalar $\phi(x)$ is a field such that $\phi'(x')=\phi(x)$.

• A one‑form $p_\alpha$ is a field that transforms the same as the gradient of a scalar:

  $$p'_\mu(x') \;=\; \frac{\partial x^\nu}{\partial x'^\mu}\,p_\nu(x).$$

  (5)

• A vector $V^\alpha$ is a field that, contracted with a one‑form, gives a scalar ($\phi = V^\alpha p_\alpha$). This forces

  $$V'^\mu \;=\; \frac{\partial x'^\mu}{\partial x^\beta}\,V^\beta.$$

  (6)

• Higher‑rank tensors are built up from these, with their transformation laws derived from the requirement that contracting with a vector or a form produces a lower‑rank tensor whose transformation law we already know.

Having defined scalar, vector, and tensor fields, one then chooses a basis $\vec e_\alpha$ at each point (any non‑singular basis will do) and writes a vector field as $\vec V = V^\alpha \vec e_\alpha$.

The contrast with the algebraic definition is that here the tensor is defined operationally by how it transforms, on a space that already has the structure of a manifold, whereas in the algebraic definition the tensor is defined constructively via tensor products with no manifold in sight.

A tensor can be defined without a manifold¶

The algebraic definition is self‑contained and needs no manifold. A tensor in the algebraic sense is simply an element of $V^{\otimes r} \otimes (V^*)^{\otimes s}$, i.e. of $T^r_s(V)$. A manifold is required only when one wants a tensor field — a tensor at each point of some space.

In that case, at each point $p$ of a manifold $M$ one attaches a tangent space $T_pM$ (a vector space), and a tensor field assigns to every $p$ an element of $T^r_s(T_pM)$. The transformation‑law definition above is really just the consistency condition for such an assignment under a change of coordinate chart on $M$.

So the picture is:

• Tensor (algebraic) = element of $V \otimes \cdots \otimes V^*$ — needs only a vector space.

• Tensor field (geometric) = a tensor at every point of a manifold — needs the manifold so the tangent spaces exist and glue together smoothly.

The two definitions in our notes are not in conflict: Differential Forms defines the algebraic object, and the differential-geometry chapter of the Theoretical Physics Reference describes how that same object behaves when it varies smoothly over a manifold.

Algebraic tensors have their own change of basis¶

Algebraic tensors come with change‑of‑basis transformations of their own, independent of any manifold. Given a vector space $V$ with basis $\{\vec e_i\}$ and a new basis $\{\vec e'_i\}$ related by an invertible matrix $M$:

components transform inversely:

• Vector: $V'^i = (M^{-1})^i{}_j\,V^j$,

• One‑form: $p'_i = M^j{}_i\,p_j$,

• Rank‑2: $A'^{ij} = (M^{-1})^i{}_k (M^{-1})^j{}_l\,A^{kl}$, and so on.

This is just linear algebra. The matrix $M$ is any invertible matrix; no manifold or coordinate system is involved.

$T_pM$ as the algebraic vector space at each point¶

At each point $p$ of a manifold $M$, the tangent space $T_pM$ is itself a vector space (of dimension $\dim M$). Tensors at $p$ are exactly the algebraic tensors over $T_pM$, built from $T_pM \otimes \cdots \otimes T_p^*M$ as above. A tensor field is a choice of such an algebraic tensor at every point.

A coordinate chart $x^\mu$ on $M$ supplies a specific basis of $T_pM$, called the coordinate basis:

When the coordinates are changed $x \to x'(x)$, the chain rule gives the new coordinate basis:

So the change‑of‑basis matrix on $T_pM$ is forced to be the Jacobian:

Plugging this $M$ into the algebraic change‑of‑basis rule of the previous section gives exactly the tensor transformation law of the manifold definition:

The unifying picture is then:

So the manifold’s transformation law is the same algebraic rule, specialized to the case where the basis comes from coordinates and the change‑of‑basis matrix is forced to be a Jacobian.

Two useful consequences follow:

• On a manifold one is free to use a non‑coordinate basis (e.g. an orthonormal tetrad / frame field). Then the change‑of‑basis matrix at each point is no longer a Jacobian — it is an arbitrary invertible matrix that may vary from point to point. The tensor transformation law still holds; it is purely algebraic at each $p$, just no longer expressible via $\partial x'/\partial x$.

• Conversely, an algebraic change of basis can always be realized locally as a coordinate transformation (any invertible matrix is the Jacobian of some local diffeomorphism), but not globally — that is a manifold‑level constraint, not an algebraic one.

The manifold’s job is therefore to (a) supply a vector space $T_pM$ at every point and (b) glue them together smoothly, so that “change of coordinates” becomes a global operation that simultaneously changes the basis on every $T_pM$ via Jacobians.

Building a tensor field from the algebraic definition¶

The previous section can be turned into a step‑by‑step recipe. Starting from the algebraic tensor, what exactly does one add to get a tensor field whose transformation laws and coordinate bases come out as consequences?

Step 1 — Algebra at a point¶

This is already done: pick a finite‑dimensional real vector space $V$ with dual $V^*$ and define

A choice of basis $\{\vec e_i\}$ of $V$ induces a dual basis $\{\tilde e^i\}$ of $V^*$ and a tensor‑product basis on $T^r_s(V)$. Any change of basis $\vec e'_i = M^j{}_i \vec e_j$ produces the algebraic transformation law on components. Nothing manifold‑shaped is needed yet.

Step 2 — A vector space at every point¶

To get a tensor field, one needs a vector space $V_p$ for every point $p$ of some set $M$. The minimum extra structure required is:

1. A smooth manifold $M$ — an atlas of charts $x: U\to\R^n$.

2. A smoothly varying family $\{V_p\}_{p\in M}$ of $n$‑dimensional real vector spaces — i.e. a rank‑$n$ vector bundle $E\to M$.

3. A smooth section — an assignment $p\mapsto T(p)\in T^r_s(V_p)$ that is smooth in $p$.

This already gives “tensor field with values in some vector bundle $E$”. But no transformation law involving Jacobians yet — a change of chart on $M$ has nothing to do with the basis of $V_p$.

Step 3 — The canonical choice $V_p = T_pM$¶

The step that couples manifold coordinates to fiber bases is the choice of $V_p$ to be the tangent space at $p$. The most transparent definition for our purposes is:

$T_pM$ is the set of equivalence classes $[(x, v)]$ where $x$ is a chart at $p$ and $v\in\R^n$, modulo

$$(x, v) \sim (x', v') \iff v'^\mu = \left.\frac{\partial x'^\mu}{\partial x^\nu}\right|_p v^\nu.$$

(13)

That is: $T_pM$ is defined by gluing copies of $\R^n$ via Jacobians. Each chart $x$ then supplies a canonical basis of $T_pM$ — namely the equivalence class of the standard basis of $\R^n$ — which one writes

By construction, two charts $x,x'$ give bases related by the Jacobian:

This is the only ingredient that links “change of coordinates on $M$” to “change of basis in $V_p$”.

Step 4 — Assemble the tensor bundle¶

From the tangent bundle $TM = \bigsqcup_p T_pM$ and cotangent bundle $T^*M$, form

where the algebra of Step 1 is applied fiber by fiber. A tensor field is a smooth section of $T^r_s M$.

Step 5 — Coordinate basis and components¶

A chart $x$ supplies the basis $\{\partial/\partial x^\mu\}$ on each fiber of $TM$ and dually $\{dx^\mu\}$ on $T^*M$, hence a tensor‑product basis on each fiber of $T^r_s M$. Components $T^{\mu_1\dots\mu_r}{}_{\nu_1\dots\nu_s}(p)$ are just the components of the algebraic tensor $T(p)$ in this basis.

Step 6 — The transformation law falls out¶

Now everything is forced. Under a chart change $x \to x'$:

• At the manifold level: the coordinates change.

• At the fiber level (Step 3): the basis of each $T_pM$ changes by the Jacobian $M^\nu{}_\mu = \partial x^\nu/\partial x'^\mu$.

• At the algebraic level (Step 1): the components transform by the algebraic change‑of‑basis rule with this specific $M$:

This is exactly the manifold transformation law: no new content, just Step 1 applied at every $p$ with $M$ chosen by Step 3.

What each ingredient contributes¶

Sanity check¶

If Step 3 were replaced by “$V_p$ = some fixed vector space $W$ independent of $TM$” (a trivial $W$‑bundle), one would still get tensor fields — but a coordinate change on $M$ would leave the fiber basis untouched. The Jacobian transformation law is not a feature of tensor fields in general; it is a feature of fields valued specifically in tensor powers of $TM$. This is why “tensors on a manifold” almost always means $T^r_s M$, and why the Jacobian shows up.

Tensors, vectors, matrices, and arrays¶

It is useful to separate two levels in the algebraic setting: the abstract object (basis‑free) and its concrete representation (basis‑dependent).

Choosing a basis is what maps one to the other:

Fix a finite‑dimensional vector space $V$ with basis $\{\vec e_i\}$. Tensors form a tower indexed by rank:

So:

• A vector is a rank‑1 tensor. Its array of components is a 1D array.

• A tensor is the general object at any rank.

• A matrix (in the strict sense) is a 2D array of numbers, i.e. the components of a rank‑2 tensor (or of a linear map) in a chosen basis.

• An array is the generic name for the multi‑dimensional grid of components at any rank.

Where “matrix” gets ambiguous¶

The word matrix is used for two different things:

1. A 2D array of numbers — purely concrete data, $A_{ij}$.

2. A linear map $f: V\to V$ — an abstract object, which is in fact a $(1,1)$‑tensor $f\in V\otimes V^*$ with components $f^i{}_j$.

These two coincide only after a basis is chosen: the “matrix of a linear map in a basis” is then the 2D array of its components.

Shape vs. type¶

The shape of an array tells you the rank, but not the type $(r,s)$. A 2D array could represent any of:

• a $(2,0)$ tensor in $V\otimes V$: components $A^{ij}$ — transforms with two Jacobians;

• a $(1,1)$ tensor in $V\otimes V^*$: components $A^i{}_j$ — one Jacobian and one inverse (the “linear map” case);

• a $(0,2)$ tensor in $V^*\otimes V^*$: components $A_{ij}$ — two inverse Jacobians (the metric is of this type).

The array $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ alone does not know which of these it is. The upper/lower index pattern carries that extra information. This is why index notation distinguishes $A^{ij}$, $A^i{}_j$, $A_{ij}$ — all three would just look like the same matrix in a programming language.

A note on the “ML tensor”¶

In machine‑learning libraries (numpy, pytorch, tensorflow), the word tensor simply means a multi‑dimensional array — only the concrete side of the table above. There is no notion of basis, of upper vs. lower indices, or of transformation law. A “rank‑3 tensor” in PyTorch is just a 3D array of numbers; it is not in general a rank‑3 tensor in the algebraic sense unless one additionally declares which axes are contravariant and which are covariant, and how they transform under change of basis.

Matrix as a linear transformation: which tensor exactly?¶

The reading of a matrix as a linear transformation identifies it with one specific tensor type: the $(1,1)$ tensor.

For a finite‑dimensional vector space $V$ there is a canonical isomorphism

So a $(1,1)$‑tensor is a linear map $V\to V$, and vice versa. No basis is needed for the correspondence.

In components: matrix multiplication is tensor contraction¶

Pick a basis $\vec e_i$ with dual $\tilde e^j$. A $(1,1)$‑tensor is

Apply it to a vector $\vec u = u^k \vec e_k$:

So the $i$‑th component of $\mathbf{A}\vec u$ is

which is exactly matrix–vector multiplication. Matrix multiplication of two linear maps $\mathbf{A}\mathbf{B}$ is the same statement:

a tensor contraction over the middle index.

Why $(1,1)$ and not $(2,0)$ or $(0,2)$?¶

A 2D array $A_{ij}$, $A^{ij}$, or $A^i{}_j$ all “look like” matrices, but they are different tensors, with different actions and different transformation laws:

Here $M^i{}_j = \partial x'^i / \partial x^j$ is the change‑of‑basis matrix. The “linear map” identification corresponds to the $(1,1)$ type, and under change of basis it transforms by the familiar similarity transformation $A' = MAM^{-1}$ — one Jacobian and one inverse, in agreement with one upper and one lower index.

By contrast, a metric or a bilinear form is $(0,2)$: it eats two vectors and gives a number. The transposes and inverses in its transformation are different from a similarity. A metric is therefore not a linear map $V\to V$, even though both can be written as $n\times n$ arrays.

Linear maps between different spaces¶

A linear map $f: V\to W$ corresponds to an element of $W\otimes V^*$. In coordinates with bases $\{\vec e_i\}$ of $V$ and $\{\vec f_a\}$ of $W$:

So a “rectangular matrix” representing $V\to W$ is a $(1,1)$ tensor with one $W$‑type upper index and one $V$‑type lower index.

Linear algebra handles several tensor types¶

Given that the $(1,1)$ reading is so canonical, one might ask whether linear algebra is simply “tensor calculus on $(1,1)$ tensors”. It is not: linear algebra actually handles several tensor types, but it usually does not distinguish them, because the standard inner product on $\R^n$ silently converts between them.

What tensor types appear in linear algebra¶

So a “matrix” in linear algebra is sometimes a $(1,1)$, sometimes a $(0,2)$, sometimes a $(2,0)$. The same 2D array of numbers stands in for very different geometric objects depending on context.

Why linear algebra can get away with conflating them¶

Two collapses happen in standard linear algebra:

1. An implicit inner product. $\R^n$ comes with the standard dot product $\langle \vec u, \vec v\rangle = \sum u^i v^i$. An inner product is a $(0,2)$ tensor $g_{ij}$ and gives a canonical isomorphism $V\cong V^*$ via $v^i \mapsto g_{ij} v^j$. So vectors and covectors stop being distinguishable: “row vector $=$ transpose of column vector”.

2. An orthonormal Cartesian basis. In such a basis $g_{ij}=\delta_{ij}$, so raising and lowering indices does nothing to the numerical components. All three of $A^{ij}$, $A^i{}_j$, $A_{ij}$ look like the same array of numbers. The type information has been thrown away.

That is why linear algebra can pretend “matrix” is a single concept: with $g = I$ in an orthonormal basis, $(2,0)$, $(1,1)$, and $(0,2)$ tensors all share the same matrix representation.

Which operations care about the type?¶

These are truly $(1,1)$‑tensorial (basis‑invariant, equivariant under the similarity transformation):

• Trace $\mathrm{tr}(A) = A^i{}_i$ — contraction of upper with lower index, requires $(1,1)$.

• Determinant of a linear map — invariant under $A\mapsto MAM^{-1}$.

• Eigenvalues and characteristic polynomial — same.

• Matrix powers $A^n$, exponential $e^A$, functional calculus — composition of linear maps.

These are $(0,2)$‑tensorial (transform as bilinear forms, $A\mapsto M^{-T}AM^{-1}$):

• Quadratic forms $\vec x^T A \vec x$.

• Inner products / Gram matrices.

• Definiteness (positive / negative definite) — only meaningful for $(0,2)$ tensors.

• Congruence, Sylvester’s law of inertia.

These are basis‑dependent or not tensorial at all without extra structure:

• Transpose $A\mapsto A^T$. Not a tensor operation in the abstract sense — there is no natural map $V\otimes V^*\to V\otimes V^*$ that does this. It only makes sense after the inner product has been used to identify $V$ with $V^*$.

• Symmetry of a matrix $A=A^T$. Meaningful only when $A$ is viewed as $(0,2)$ or $(2,0)$; ill‑defined as a property of a $(1,1)$‑tensor (a linear map).

• Orthogonal matrices are defined by preserving the inner product, so they presuppose a $(0,2)$ metric.

Reading a linear algebra theorem, one can almost always tell which tensor type it secretly assumes by which of $\{\text{trace, determinant, eigenvalues}\}$ vs $\{\text{symmetry, quadratic form, positive definite}\}$ vs $\{\text{orthogonal, transpose}\}$ it uses.

Beyond rank 2¶

Linear algebra essentially stops at rank 2. Once one needs rank‑3 or higher objects (e.g. the Riemann curvature tensor $R^\mu{}_{\nu\alpha\beta}$, the structure constants of a Lie algebra $c^k{}_{ij}$, the elasticity tensor $C_{ijkl}$, the Levi‑Civita symbol $\varepsilon_{ijk}$), there is no good “matrix” language for them, and one must use tensor / index notation. That is the regime where the algebraic tensor framework becomes essential and linear algebra runs out.

Refining the picture of “matrix”¶

It is now precise to say that matrix admits two definitions:

• The concrete definition: a 2D array of numbers, with no geometric meaning.

• The abstract definition: the components of a rank‑2 tensor in a basis, which in linear algebra is silently allowed to be any of $(1,1)$, $(2,0)$, or $(0,2)$.

The collapse “all three look identical” requires two ingredients:

1. An inner product (a $(0,2)$ metric $g$) to identify $V\cong V^*$ — this is what lets one raise / lower indices at all.

2. An orthonormal basis so that $g_{ij}=\delta_{ij}$ — this is what makes the raising / lowering numerically a no‑op.

Without orthonormality, even with an inner product, the components of $A^{ij}$, $A^i{}_j$, $A_{ij}$ differ; raising an index multiplies by $g^{ij} \neq \delta^{ij}$. And the $(1,1)$ “linear map” reading is the only one that needs no metric at all — it is self‑contained via the canonical isomorphism $V\otimes V^* \cong \mathrm{Hom}(V,V)$. Only the $(2,0)$ and $(0,2)$ readings need the metric to be related to it.

“Vector” also has multiple meanings¶

The same story plays out, one rank lower, for the word vector.

Senses of “vector”¶

The same conflation as for “matrix”¶

In standard Euclidean $\R^n$ with the standard dot product and orthonormal Cartesian basis, the components of the $(1,0)$ vector $v^i$ and of the $(0,1)$ form $v_i$ are numerically identical: $v_i = \delta_{ij} v^j = v^i$. So:

• Vector and covector collapse into one “1D array of numbers”.

• Column vector and row vector become just transposes of each other.

The collapse requires both (i) an inner product to identify $V\cong V^*$ and (ii) an orthonormal basis to make the identification numerically trivial — exactly the same story as for matrices. In a non‑orthogonal basis, even with the standard inner product, $v^i \neq v_i$ as numbers; that is why differential geometry insists on the distinction.

What gives away which sense is meant¶

• “Vector space of polynomials of degree $\leq 5$” — sense 2 (abstract).

• “The gradient $\nabla f$ is a vector” — it is really a $(0,1)$ form, not a $(1,0)$ vector; the conflation is a metric‑induced abuse.

• “Velocity $\vec v$ is a vector” — $(1,0)$ tensor, genuinely contravariant.

• “Angular velocity $\vec\omega$ is a vector” — pseudovector; really an antisymmetric $(0,2)$ tensor in disguise (in 3D, via the Hodge star).

• “Tangent vector to a curve” — element of $T_pM$, i.e. $(1,0)$ at a point on a manifold.

• v = np.array([1,2,3]) — 1D array; no geometric content unless one is supplied.

Parallel with “matrix”¶

So the situation is structurally identical, one rank lower.

Pseudotensors at every rank and in every dimension¶

Pseudovectors are not a quirk of 3D vectors — they are a special case of pseudotensors, which exist at every rank and in every dimension. The “pseudo” prefix is a parity twist that is orthogonal to rank and dimension; a “matrix” can be a pseudotensor just as well as a vector can.

Definition¶

A pseudotensor transforms like a tensor under orientation‑preserving changes of basis but picks up an extra sign $\operatorname{sgn}(\det M)$ under orientation‑reversing ones:

The $\operatorname{sgn}(\det M)$ factor is the only difference from a genuine tensor. There are therefore parallel hierarchies:

So a “matrix” can absolutely be a pseudotensor, and this is dimension‑independent.

2D examples¶

• 2D Levi‑Civita symbol $\varepsilon_{ij}$ with $\varepsilon_{12} = -\varepsilon_{21} = 1$. This $2\times 2$ array is literally a pseudomatrix — an antisymmetric rank‑2 pseudotensor of type $(0,2)$.

• 2D “cross product” $u\times v \equiv \varepsilon_{ij} u^i v^j$ is a pseudoscalar (signed area). Under a reflection of the $xy$‑plane the geometric area is unchanged but its sign flips, which is exactly the pseudoscalar behavior.

• 2D curl $(\partial_x v_y - \partial_y v_x) = \varepsilon^{ij}\partial_i v_j$ is a pseudoscalar.

• The outer product of a regular vector with a pseudovector gives a pseudomatrix.

3D: how pseudovectors are “really” pseudomatrices¶

This makes precise a remark in Differential Forms. In 3D, an antisymmetric rank‑2 tensor $A_{ij}$ has 3 independent components, and one can identify them with a vector via

Because $\varepsilon_{ijk}$ is a pseudotensor:

• If $A_{ij}$ is a genuine (antisymmetric) tensor, then $B^k$ is a pseudovector (one $\varepsilon$ flips parity once).

• If $A_{ij}$ is itself a pseudotensor, then $B^k$ is a genuine vector (two pseudos cancel).

This is why the magnetic field $\vec B$ and the angular velocity $\vec\omega$ are pseudovectors: they are really antisymmetric rank‑2 tensors in disguise, and the disguise is performed by $\varepsilon_{ijk}$.

The deeper unification¶

A pseudotensor is just a tensor on an unoriented manifold, or equivalently a section of the bundle

where $\mathrm{or}(M)$ is the orientation line bundle (a $\mathbb{Z}_2$‑twist). Choosing an orientation trivializes this bundle and lets one treat pseudotensors as ordinary tensors — until the orientation is changed, at which point the sign flip becomes visible.

Equivalently, pseudotensors are tensor densities of weight 0 with a sign character: they differ from ordinary tensors only in carrying the sign representation of $GL(n,\R) \to \{\pm 1\}$ given by $\operatorname{sgn}(\det)$.

Tensor densities and the determinant of $A_{ij}$¶

The same language also gives the clean way to say what the determinant of a covariant rank-2 tensor means. The important point is that $\det(A_{ij})$ is not naturally a scalar. It is naturally a density-like object.

Relative invariants¶

Let $J\in GL(n,\R)$ be a change-of-basis matrix, and let $J*A$ denote the induced action on whatever object $A$ is. A scalar-valued expression $F(A)$ is a relative invariant if

for some factor $\chi(J)$ depending only on $J$, not on $A$. If $F$ is not identically zero, then $\chi$ is forced to be a group homomorphism. Indeed, for some $A$ with $F(A)\ne0$,

while associativity of the group action gives

Thus

The determinant notes show the corresponding principle for matrix homomorphisms: a multiplicative character of $GL(n)$ factors through the determinant. Over $\R$, with the usual regularity assumptions used for Cauchy-type equations (continuity, or merely measurability), the characters are

where only the parity of $k$ matters. This is the tensor-density classification. For objects with free tensor indices, the table records the extra density factor multiplying the ordinary tensor index transformation.

Here $\lvert g\rvert$ abbreviates $\lvert\det g_{ij}\rvert$. For example, the lower Levi-Civita symbol row means

so it is not an ordinary $(0,n)$ tensor; it is a $(0,n)$ tensor density with $(w,k)=(-1,1)$. Similarly, the upper Levi-Civita symbol has $(w,k)=(1,1)$.