Where is the magical sign change under change of basis ? Not pseudotensor?

Question

I'm sorry for the long post but the this subject is confusing to me.

Context: On one hand wiki talks about pseudovectors as if they are maps $\Phi:V^k \to V$ on the physical vector space with the property $\Phi(R x_1,…,R x_k) = (\det R)R \Phi(x_1,…,x_k)$, where $R$ is a linear isometry. This is a property fulfilled by the cross product.

On the other hand there are pseudotensors, whose components, when changing basis (supposedly a passive transformation), change as those of a tensor but get multiplied by the sign of the detreminant of that transformation, i.e. $$ {A^{i_1'\cdots i_p'}}_{j_1'\cdots j_q'} = (-1)^\alpha \alpha^{i_1'}_{k_1'}\cdots\alpha^{i_p'}_{k_p} \alpha^{l_1}_{j_1'}\cdots\alpha^{l_q}_{j_p'} {A^{k_1\cdots k_p}}_{l_1\cdots l_q} $$ where ${A^{i_1'\cdots i_p'}}_{j_1'\cdots j_q'}$ and ${A^{k_1\cdots k_p}}_{l_1\cdots l_q}$ are the new and old components of the "tensor" $A$ respectively and $(-1)^\alpha$ is the sign of the determinant of the transformation between the new and old bases.

So I thought that map of the cross product, if expressed as a "tensor", should exhibit this behaviour of pseudotensors, since it is a pseudovector, an instance of a pseudotensor.

My work: Let $V$ be a real 3-dimensional inner product space. The cross product $\kappa:V \times V \to V$ can be defined in many ways, e.g. $\star \circ \wedge$. For this definition I tried to find the tensor representation and I obtained in the arbitrary basis $E=(e_1,e_2,e_3)$ the expression (derivation below) $$ \kappa = \det (T^U_E) \epsilon_{\vec{rp},m} g^{mn} e_n \otimes (e^r \wedge e^p) = \det (T^U_E) \epsilon_{rpm} g^{mn} e_n \otimes e^r \otimes e^p = {\kappa^n}_{rp}\ e_n \otimes e^r \otimes e^p $$ where $T^U_E:V \to V$ is the transformation from an oriented orthonormal basis $U$ to the arbitrary basis $E$, $g^{mn}:=\langle e^m_\flat, e^n_\flat\rangle$ are components of the dual metric tensor, i.e. the inner product on $V^*$ induced from that on $V$ via the musical isomorphism, and the arrow over $rp$ indicates increasing indices. I hope also this formula confirmed correct.

Now I tried to verify whether under change of basis $T^E_{E'}:E \mapsto E'$ the components ${\kappa^n}_{rp}$ of this "tensor" transform as those of a pseudotensor. The transformation for the basis vectors $e_{i'} = \alpha^{j}_{i'} e_j$, therefore for the dual basis vectors it is $e^{i'} = \alpha^{i'}_{j} e^{j'}$, where $\alpha^{i'}_{j}\alpha^{j}_{k'} = \delta^{i'}_{k'}$ and $\alpha^{i}_{j'}\alpha^{j'}_{k} = \delta^{i}_{k}$. Therefore $$ e_n \otimes e^r \otimes e^p = \alpha_n^{i'}\alpha_{j'}^{r}\alpha_{k'}^{p}\ e_{i'} \otimes e^{j'} \otimes e^{k'} $$ this suggests the transformation law should be \begin{align*} {\kappa^n}_{rp}\ \alpha_n^{i'}\alpha_{j'}^{r}\alpha_{k'}^{p} &= \det (T^U_E) \epsilon_{rpm} g^{mn} \alpha_n^{i'}\alpha_{j'}^{r}\alpha_{k'}^{p} \\ &= \det (T^U_E) \epsilon_{rpm} g^{l's'}\alpha_{l'}^m \alpha_{s'}^n \alpha_n^{i'}\alpha_{j'}^{r}\alpha_{k'}^{p} \\ &= \det (T^U_E) \epsilon_{rpm} g^{l's'}\alpha_{l'}^m \delta_{s'}^{i'} \alpha_{j'}^{r}\alpha_{k'}^{p} \\ &= \det (T^U_E) \epsilon_{rpm} g^{l'i'}\alpha_{l'}^m \alpha_{j'}^{r}\alpha_{k'}^{p} \\ &= \det (T^U_E) \det(T^E_{E'})\epsilon_{j'k'l'} g^{l'i'} \\ &= \det (T^U_E T^E_{E'})\epsilon_{j'k'l'} g^{l'i'} \\ &= \det (T^U_{E'})\epsilon_{j'k'l'} g^{l'i'} = {\kappa^{i'}}_{j'k'} \\ \end{align*} where $\det(T^E_{E'}) \epsilon_{j'k'l'} = \epsilon_{rpm}\alpha_{l'}^m \alpha_{j'}^{r}\alpha_{k'}^{p}$.

Could anyone tell me where the mistake is if there is any and why it doesn't transform as a pseudotensor ?

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Derivation of the tensor expression of the cross product from different definition

Let's define the cross product of two vectors $x,y$ as $$ \kappa(x,y) = \left(i(y) i(x) \text{d}vol^3\right)_\flat $$ where $i(x)$ is the interior product (insertion operator or whatever it is called) and $\text{d}vol^3 = (\pm)\sqrt{\det g_*}e^1\wedge e^2\wedge e^3 = \det(T^U_E) e^1\wedge e^2\wedge e^3$ is the oriented volume form and $U$ is the oriented orthonormal basis, $T^U_E:V \to V$ the transformation from $U$ to $E$, and $g_*:=(g_{ij})_{ij}, g_{ij} := \langle e_i,e_j\rangle$ the matrix of the metric tensor (of the inner product $g\equiv \langle \cdot,\cdot\rangle$) on $V$ (with indices as subscripts) in the basis $E$. Finally the map $\flat:V^* \to V$ is a musical isomorphism that identifies the space $V$ and its dual $V^*$, its inverse $\sharp:V \to V^*, x \mapsto \langle x, \cdot \rangle$. Then \begin{align} (i(y) i(x) \text{d}vol^3)_\flat &= \left(i(y) i(x) \det(T^U_E) e^1\wedge e^2\wedge e^3\right)_\flat \\ \text{Summation over (2,1)-shuffles} &= \det(T^U_E) \left(\sum_{(\vec{rp},m) \in S(2,1)} \epsilon_{\vec{rp},m} (e^r\wedge e^p)(x,y)\ e^m \right)_\flat \\ \text{now omit the summation sign} &= \det(T^U_E) \epsilon_{\vec{rp},m} (e^r\wedge e^p)(x,y)\ (e^m)_\flat \\ &= \det(T^U_E) \epsilon_{\vec{rp},m} (e^r\wedge e^p)(x,y)\ g^{m n} e_n \\ &= \det(T^U_E) \epsilon_{\vec{rp},m} g^{m n} e_n \otimes (e^r\wedge e^p)\ (x,y)\\ \end{align} where the formula for the interior product is from this post.

Answer 1

Let $g$ be an inner product on the $n$ dimensional vector space $V$ (not necessarily positive definite). Suppose that an orientation has been chosen on $V$. Then there is an $n$-form $\mu_g\in\Lambda^n(V^\ast)$ specified uniquely by $$ \mu_g(e_1,\dots,e_n)=1 $$for any positively oriented orthonormal basis. Although pseudo $n$-forms and "densities" are equivalent, let's use a more adequate definition of "pseudo-ness".

A pseudo-scalar on $V$ is a map $\mathcal B(V)\rightarrow\mathbb R$ where $\mathcal B$ is the set of all ordered bases of $V$ such that for any $A\in\mathrm{GL}(n,\mathbb R)$, we have $$ f(eA)=(\mathrm{sgn}\,\det A)f(e), $$ i.e. a pseudo-scalar associates a number to each basis of $V$ such that if the basis is changed, the number gets multiplied by the sign of determinant of the basis change. Let $\mathrm{PSc}(V)$ the set set (vector space) of pseudo-scalars on $V$.

A pseudo $n$-form on $V$ is then an element of $\mathrm{PSc}(V)\otimes\Lambda^n(V^\ast)$, and we also have a canonical pseudo $n$-form given by $m_g(e_1,\dots,e_n)=1$ for any orthonormal basis. Here a pseudo $n$-form is an alternating multilinear map from $n$ vectors to a pseudo-scalar.

This is indeed a pseudo $n$-form since if $A\in \mathrm O(g)$ is a (generalized) orthogonal transformation, and $\bar e=eA$, then $m_g(\bar e_1,\dots,\bar e_n)=(\mathrm{sgn}\,\det A)^2m_g(e_1,\dots,e_n)=m_g(e_1,\dots,e_n)$, where the sign factor is squared because it gets a sign factor from the $n$-form part of $m_g$ and gets another sign factor from the pseudo-scalar part of of $m_g$.

Stated differently, the orientation of a vector space is a pseudo-scalar, namely $\sigma(e)=\pm 1$ where the + sign is chosen for a positively oriented basis and the - sign for a negatively orented basis. Then simply define $$ m_g:=\sigma\otimes\mu_g $$ which turns out to be independent of the chosen orientation.

Now if $\varepsilon_{i_1...i_n}$ is the algebraic Levi-Civita symbol, then for any ONB $e$ we have $$ \mu_g(e_{i_1},\dots,e_{i_n})=\pm\varepsilon_{i_1...i_n}, $$ where the sign depends on the orientation of the basis, and $$ m_g(e_{i_1},\dots,e_{i_n})=\varepsilon_{i_1...i_n}, $$where there is now no orientation-dependent sign factor.

Now let $n=\dim V=3$, and we define two different cross products: $$ u\times^\prime v:=\mu_g(-,u,v) \\ u\times v:=m_g(-,u,v), $$ where I have left a metric duality implicit (i.e. these cross products are covector-valued, but then "raise the index", so to speak, to get the usual vector-valued cross product).

The first "primed" cross product is a tensor. The second "unprimed" cross product is a pseudo-tensor.

As it happens, in most mathematical-oriented literature, involving Hodge stars and all that, the "$\times^\prime$" definition of the cross product is used, and is thus a tensor.

In most physics texts, especially field theory texts, the cross product is defined through the Levi-Civita symbol and one assumes the Levi-Civita symbol to be the same in all (Cartesian) coordinate systems. Therefore, this Levi-Civita symbol is the representation of the pseudo $n$-form $m_g$ and not the $n$-form $\mu_g$. Hence, this cross product is the "unprimed" $\times$ cross product, which is a pseudo-tensor.

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Edit:

With regards to the comments below the other answer, I want to clarify the relationship between coordinate-based definitions of various geometric objects and more map-based definitions.

Let us define the jet group of order $s$, denoted as $G^s(m)$. First let $\mathrm{Diff}_0(m)=\{\varphi:U\rightarrow\mathbb R^m:\ U\subseteq\mathbb R^m\text{ is open}, 0\in U, \varphi(0)=0, D\varphi(0)\text{ is an isomorphism}\}$ be the set of all local diffeomorphisms of $\mathbb R^m$ into itself, defined on some neighborhood of $0$, transforming $0$ into $0$.

Then introduce the equivalence relation $\sim_s$ on $\mathrm{Diff}_0(m)$, whereby $\varphi\sim_s\varphi^\prime$ if and only if$$ D^k\varphi(0)=D^k\varphi^\prime(0),\quad 1\le k\le s, $$ i.e. two such local diffeomorphisms are $s$-equivalent if and only if all their derivatives at $0$ up to and including order $s$ agree. Let $G^s(m):=\mathrm{Diff}_0(m)/\sim_s$ be the set of equivalence classes. Note that each element of $G^s(m)$ has a canonical representation as a truncated Taylor polynomials, eg. if $\varphi\in G^s(m)$ then $$ \varphi^i(x)=a^i_j x^j+\frac{1}{2}a^i_{jk}x^j x^k+\dots+\frac{1}{s!}a^i_{j_1...j_s}x^{j_1}\dots x^{j_s} $$ subject only to the condition that $\det{(a^i_j)}\neq 0$.

The group law in $G^s(m)$ is given by truncated composition, i.e. for $\varphi_1,\varphi_2\in G^s(m)$, we have $$ \varphi_1\varphi_2:=(\varphi_1\circ\varphi_2)_s, $$ where $(-)_s$ means that the resulting polynomial of degree $2s$ is cut off at degree $s$.

Obviously $G^1(m)\equiv \mathrm{GL}(m,\mathbb R)$.

Now let $M$ be a smooth $m$ dimensional manifold. A geometric object of type $(\lambda,F)$ is described by the data $(\lambda,F)$, where $F$ is a smooth manifold and $\lambda:G^s(m)\times F\rightarrow F$ is a smooth left action of the jet group $G^s(m)$ on $F$. This case, the geometric object is said to be of order $s$.

More specifically, a geometric object $\Phi$ of type $(\lambda,F)$ at $p\in M$ is a map $\Phi:\mathcal A_p\rightarrow F$ (where $\mathcal A_p$ is the set of all coordinate charts of $M$ that contain the point $p$) obeying a certain equivariance condition. I need to set up some conventions and notations to describe it. First of all, let $\Phi[x]$ denote the value of the geometric object on the chart $(U,x)$ (the domain is unimportant as by assumption all domains contain $p\in M$).

Now given any pair of charts $(U,x)$ and $(\bar U,\bar x)$ both containing $p$, there is a unique element $g^s(\bar x,x)\in G^s(m)$ determined by the truncated Taylor expansion of the coordinate transformation function $\bar x(x)$ translated so that $\bar x(p)=x(p)=0$ (this translation is needed because we have defined the jet groups through diffeomorphisms transforming $0$ into $0$).

The desired equivariance property is then $$ \Phi[\bar x]=\lambda\left(g^s(\bar x,x),\Phi[x]\right). $$

Since this is a bit abstract, let's unwrap the definition:

• A geometrical object of type $(\lambda,F)$ at $p\in M$ associates to each coordinate system $(U,x)$ an element $\Phi[x]\in F$ of $F$ which we may view as the "components" of $\Phi$ in the coordinate system $(U,x)$ (in practice, $F=\mathbb R^n$ for some $n\in\mathbb N$ usually, hence the name "components").
• When the coordinate system is changed, i.e. we go from $(U,x)$ to $(\bar U,\bar x)$, the "components" $\Phi[\bar x]$ in the new system are related to the "components" $\Phi[x]$ in the previous system via a group action $\Phi[\bar x]=g^s(\bar x,x)\cdot\Phi[x]$ (here I abbreviated the group action from $\lambda(-,\ast)$ to $-\cdot\ast$) of the jet group of order $s$.
• The jet group element $g^s(\bar x,x)$ is just determined by the values$$ \frac{\partial \bar x^i}{\partial x^j}(p),\ \frac{\partial^2\bar x^i}{\partial x^{j_1}\partial x^{j_2}}(p),\dots,\frac{\partial^s \bar x^i}{\partial x^{j_1}\dots\partial x^{j_s}}(p) $$ with the group composition being the truncated composition of Taylor polynomials.
• Hence, a geometric object of type $(\lambda, F)$ with $\lambda$ being the action of the jet group of order $s$ is a geometric object with components in $F$ and whose transformation law involves the $s$-fold derivatives of the coordinate transformation functions.

Every tensor is a geometric object of order $1$ with $F=\mathbb R^k\otimes\mathbb R^{\ast l}$ for a type $(k,l)$ tensor and the action of the jet group (remember that $G^1(m)=\mathrm{GL}(m,\mathbb R)$) the usual linear representation of $G^1(m)$ on the tensor product.

A linear connection is a geometric object of order $2$. After all its transformation law includes second derivatives. Exercise: Determine $F$ and $\lambda$ for a linear connection!

Now let $LM\rightarrow M$ be the linear frame bundle of $M$, i.e. it is a principal $G^1(m)$-bundle such that each point of $LM$ is a frame $e\in LM$, i.e. a basis of $T_pM$ for some $p\in M$ with projection $\pi:LM\rightarrow M$ transforming a basis $e$ of $T_pM$ into $p$. The principal bundle structure is given by $eA=e_i a^i_{\ j}$, where $A=(a^i_{\ j})$ is an invertible matrix.

The idea is that if $\bar e,e\in L_pM$ are both bases at $p$, then both $\bar e$ and $e$ are always coordinate bases, since the difference between a holonomic and a non-holonomic frame is a differential condition, so a frame can only be non-holonomic if we at least know the first derivative of the frame at a point.

Therefore, if $\Phi$ is a geometric object of type $(\lambda,F)$, where $\lambda$ is a left action of the the jet group $G^1(m)$, then this object can be described as a map $\Phi:LM\rightarrow F$ obeying an equivariance condition $$ \Phi(eA)=A^{-1}\cdot\Phi(e) $$(again, "$\cdot$" denotes the left action $\lambda$) [actually, this describes a field of geometrical objects; for a single geometric object at $p\in M$, replace $LM$ with $L_pM$.].

Note that for higher order geometric objects, this description does not perfectly work in that a geometric object of order $s\ge 2$ can be described as a map on the "order $s$ frame bundle" $L^sM$, but this "higher order frame bundle" now only contains holonomic frames, so at this point we must make a difference between objects that behave in a certain way with respect to bases and objects which behave in a certain way with respect to coordinate systems.

But for geometric objects of order $1$, this distinction does not exist.

To actually get to the topic at hand, i.e. "pseudo-things", let's say that a given source describes a pseudo scalar at $p\in M$ as a rule which associates to each coordinate system $(U,x)$ a number $a\in\mathbb R$ such that when the coordinate system is changed to $(\bar U,\bar x)$, the number changes as $$ \bar a=\mathrm{sgn}\det\left(\frac{\partial\bar x}{\partial x}(p)\right)a. $$ Then the pseudo-scalar is a geometrical object of order $1$, therefore, there is automatically a description as follows. A pseudo-scalar $f$ at $p\in M$ is a function $f:L_pM\rightarrow \mathbb R$ such that $$ f(eA)=\mathrm{sgn}\det(A^{-1})f(e) $$ whenever $A\in G^1(m)$ is an invertible matrix describing a change of basis at $p$.

Note that in the notation of my original answer, $T_pM=V$, $\mathcal B(V)=L_pM$ can we get back the original definition (that here we have $A^{-1}$ instead of $A$ does not matter as the sign is the same for both).

You can use this technique to convert any geometric object of order $1$ defined in terms of a coordinate system to one defined as a function on bases.

Answer 2

After typing up my answer, I realised that Bence had already answered with many similarities to what I had the intention to write. Hence I present my answer trying to conform to Bence's notation, so that it's less painful to read both side to side. $\def\sgn{\mathrm{sgn}}$

Pseudoscalars

Scalars can be defined without the need of a coordinate system. For example, the scalar $1$ has always the same value. Moreover, every scalar $a$ is a multiple of $1$, since $a=1a$. On the other hand, there are other objects called pseudoscalars which are almost like scalars, except they have a coordinate representation whose sign depends on the orientation of the basis. If $e=(e_1,\dots,e_n)$ is a basis for $V$, then you can construct a pseudoscalar $$\sgn(e)$$ This object only has a magnitude, but not a sign. What I mean is that if $b=(b_1,\dots,b_n)$ is another basis, which is related to $e$ by $b_i=A^j{}_ie_j$ (denote this by $b=eA$) then $\sgn(b)=\sgn(eA)=\sgn(A)\sgn(e)$. Hence, for example, if $b=(e_1,\dots,e_{n-1},-e_{n})$, you have $$\sgn(b) = -\sgn(e).$$ This pseudoscalar is like $1$ in the sense that any pseudoscalar $\sigma$ can be expressed as a multiple of $\sgn(e)$ as $$\sigma = \sigma_{(e)}\sgn(e)$$ where $\sigma_{(e)}\in F$ is a number, called the component of $\sigma$ with respect to the frame $e$. Now we will obtain the expression for $\sigma$ with respect to another frame $b$. If $b_i=A^j{}_ie_j$, as before, then you have $$\sigma_{(b)}\sgn(b)=\sigma=\sigma_{(e)}\sgn(e)=\sigma_{(e)}\sgn(bA^{-1})=\sigma_{(e)}\sgn(A^{-1})\sgn(b)=\sgn(A)\sigma_{(e)}\sgn(b),$$ so that $$\sigma_{(b)}=\sgn(A)\sigma_{(e)}.$$ This is the behaviour of the component of the pseudoscalar $\sigma$ under a change of basis, but observe that $\sigma$ itself is an invariant object, the same way a tensor is an invariant object even if its components with respect to different coordinate systems are different. Also note that a scalar multiple of a pseudoscalar is again a pseudoscalar (in particular, pseudoscalars form a vector space over the base field), while the product of two pseudoscalars is a scalar. (Notice: what Bence called a pseudoscalar is what I call the component of a pseudoscalar).

Pseudotensors

Now, in physics

a $(p,q)$-tensor is a multilinear map that "eats $q$ vectors, $p$ covectors and spits out a scalar"

right? You are already familiar with the representation of a tensor $T$, say of rank $(0,2)$, with respect to a basis $e=(e_1,\dots,e_n)$ as a sum $$T_{ij,(e)}e^i\otimes e^j$$ where $T_{ij,(e)}$ are the components of $T$ with respect to $e$. I used the sum convention and of course $e^i$ are the dual vectors of $e_i$.

Now

a pseudo-$(p,q)$-tensor is a multilinear map that "eats $q$ vectors, $p$ covectors and spits out a pseudoscalar"

Similarly, the coordinate representation of a pseudotensor $P$ of rank, say, $(0,2)$, is $$P_{ij,(e)}e^i\otimes e^j \otimes\sgn(e),$$ where you insist that the basis that goes inside $\sgn(e)$ is the same one over which you are taking the linear combination. Now, if $b=eA$ is another basis you have $b_i=A^j{}_ie_j$, so that $e^j=A^j{}_ib^i$ and so the components of $P$ in the frame $b$ are $$\begin{align} P_{ij,(b)}b^i\otimes b^j \otimes\sgn(b) &=P \\ &=P_{kl,(e)}e^k\otimes e^l \otimes\sgn(e) \\ &=P_{kl,(e)}A^k{}_ib^i\otimes A^l{}_jb^j \otimes\sgn(bA^{-1}) \\ &=\sgn(A)A^k{}_iA^l{}_jP_{kl,(e)}b^i\otimes b^j \otimes\sgn(b) \\ \end{align}$$ so that the components of $P$ obey the transformation rule $$P_{ij,(b)}=\sgn(A)A^k{}_iA^l{}_jP_{kl,(e)}$$ as you were expecting.

Example

Take an $n$-dimensional vector space $V$ and fix an $n$-form $\mu\neq 0$. Then you can define the volume of a set of $n$-vectors $(u_1,\dots,u_n)$ relative to $\mu$ as $$\mu(u_1,\dots,u_n)$$ Then $\mu$ induces a pseudo-$n$-form $m$ as follows: take any basis $e=(e_1,\dots,e_n)$ of $V$ and define $$m=\mu\otimes\sgn(e).$$ I.e. $$m(u_1,\dots,u_n)=\mu(u_1,\dots,u_n)\sgn(e).$$ While $\mu$ is an $n$-form, $m$ is a pseudo-$n$-form, so its components transform as follows: whenever $b=eA$, we get $$\begin{align} m_{i_1\dots i_n,(b)} &= m(b_{i_1},\dots,b_{i_n})\sgn(b) \\ &= \mu(b_{i_1},\dots,b_{i_n})\sgn(e)\sgn(b) \\ &= \mu(A^{j_1}{}_{i_1}e_{j_1},\dots,A^{j_n}{}_{i_n}e_{j_n})\sgn(e)\sgn(eA) \\ &= (\det A)\mu(e_{i_1},\dots,e_{i_n})\sgn(A) \\ &= |\det A|\mu(e_{i_1},\dots,e_{i_n}) \\ &= |\det A|\mu(e_{i_1},\dots,e_{i_n})\sgn(e)\sgn(e) \\ &= |\det A|m(e_{i_1},\dots,e_{i_n})\sgn(e) \\ &= |\det A|m_{i_1\dots i_n,(e)} \end{align}$$ where the first and last equalities can be seen applying both sides of $$\begin{align} m_{j_1\dots j_n,(b)} b^{j_1}\otimes\dots\otimes b^{j_n}\otimes\sgn(b) &= m \\ m_{j_1\dots j_n,(e)} e^{j_1}\otimes\dots\otimes e^{j_n}\otimes\sgn(e) &= m \end{align}$$ to $(b_{i_1},\dots,b_{i_n})$ and $(e_{i_1},\dots,e_{i_n})$, respectively.

Now, what is the issue with the cross product? For a simpler case consider the "scalar cross product" defined on a two-dimensional space. For example, take any vector space of dimension $2$ and fix a basis $e_1,e_2$. Then define a $2$-form $$\mu = e^1\otimes e^2-e^2\otimes e^1$$ and a pseudo-$2$-form $m=\mu\otimes\sgn(e)$. As Bence said, these are two notions of (scalar) cross product. For any vectors $u=u^1e_1+u^2e_2$ and $v=v^1e_1+v^2e_2$, you have $$\mu(u,v)=u^1v^2-u^2v^1,$$ $$m(u,v)=\mu(u,v)\sgn(e)=(u^1v^2-u^2v^1)\sgn(e).$$ Note that, while $\mu(u,v)$ is a mere number, which can be positive or negative, the quantity $m(u,v)$ is a pseudoscalar, so you can always find a basis such that the component of $m(u,v)$ is positive. For example, if $u=e_2$ and $v=e_1$, you have $$\mu(u,v)=-1$$ but $$m(u,v)=-\sgn(e)=\sgn(b)$$ where $b$ is any basis with opposite orientation to $(e_1,e_2)$. In that sense, $\mu(u,v)$ can be interpreted as a signed area, while, $m(u,v)$ can be thought of as an unsigned area.

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Edit: Another view of pseudoscalars

My initial idea was to define a pseudoscalar as "an element of the representation space of the sign representation of $GL(V)$", which as you suggested is akin to saying that a tensor is "an element of the representation space of a tensorial representation of $GL(V)$", i.e. the mathematician's version of the infamous "something that transforms like a tensor".

As you say, pseudoscalars can indeed be represented by functions $$\sigma : \Lambda^nV\setminus\{0\}\to\Bbb R$$ with the property $$\sigma(t\omega) = \sgn(t)\sigma(\omega)$$ for all $\omega\in\Lambda^nV$, $\omega\neq 0$, $t\in\Bbb R$, $t\neq 0$. Call the space of these functions $\mathrm{PSc}(V)$. This is similar to how the $(p,q)$-tensors can be represented as multilinear maps $\mathrm{Mult}^{p+q}((V^*)^p,V^q;\Bbb R)$. At the time of writing my original post, I hadn't quite realized this, so I apologize for the confusion. With this representation, given a basis $e$ of $V$, then $\sgn(e)$ is the unique pseudoscalar with $$\sgn(e)(e_1\wedge\dots\wedge e_n)=1,$$ as you suggested in the comments. Then $\sgn(e)$ is a basis of the space of pseudoscalars: any pseudoscalar $\sigma\in\mathrm{PSc}(V)$ can be written uniquely as $$\sigma = \sigma(e_1\wedge\dots\wedge e_n)\sgn(e).$$

About your last comment, you are right, components of pseudotensors are scalars. And yes, the arguments of $m=\mu\otimes\sigma$ are independent of each other. The result of $m$ acting on a pair of vectors is a pseudoscalar-valued $1$-form. I.e. a pseudo-$1$-form: $$m(\_,u,v) = \mu(\_,u,v)\otimes\sigma.$$ If you use the sharp isomorphism, you will get Bence's pseudovector valued cross product: $$u\times v = m(\_,u,v)^\sharp = \mu(\_,u,v)^\sharp\otimes\sigma.$$

About the literature, for the construction of associated bundles I suggested

• Natural operations in differential geometry, chapter III, 10.7. - Ivan Kolar, Peter W. Michor, Jan Slovak
• Foundations of differential geometry, by Kobayashi and Nomizu, volume I, page 53. They denote associated bundles by $E(M,F,G,P)$. See the examples 5.2, 5.3, 5.4 etc.

The ones I found specifically about pseudo forms are

• Theodore Frankel - The geometry of physics. Section 2.8e.
• William L. Burke - Div, grad, curl are dead
• William L. Burke - Twisted forms: Twisted differential forms as they should be. You can find it here.

About your comment on whether the components of $m$ should be scalars: yes, they should. I guess the equation $$m_{i_1\dots i_n,(b)} = m(b_{i_1},\dots,b_{i_n})\sgn(b)$$ isn't very clear about that, but $m(b_{i_1},\dots,b_{i_n})\sgn(b)$ is a product of two pseudoscalars: $m(b_{i_1},\dots,b_{i_n})$ and $\sgn(b)$, and hence it is a true scalar. Another way to derive the transformation law of the components of $m=\mu\otimes\sgn(e)$, now interpreting pseudoscalars as functions $\Lambda^nV\setminus\{0\}\to\Bbb R$, is $$\begin{align} m_{i_1\dots i_n,(b)} &= m(b_{i_1},\dots,b_{i_n})(b_1\wedge\dots\wedge b_n) \\ &= \mu(b_{i_1},\dots,b_{i_n})\sgn(e)(b_1\wedge\dots\wedge b_n) \\ &= \mu(A^{j_1}{}_{i_1}e_{j_1},\dots,A^{j_n}{}_{i_n}e_{j_n})\sgn(e)((\det A)e_1\wedge\dots\wedge e_n) \\ &= |\det A|\mu(e_{i_1},\dots,e_{i_n}) \sgn(e)(e_1\wedge\dots\wedge e_n)\\ &= |\det A|m_{i_1\dots i_n,(e)} \end{align}$$