Linear and tensor methods repair

Component calculations become reliable only when every array has an invariant meaning. This focused review rebuilds the distinction between a vector and its coordinates, between a covector and a row of numbers, and between a linear map and the matrix used to represent it. The payoff in QFT is immediate: index placement, adjoints, kinetic terms, and changes of field basis stop being rules to memorize and become consequences of types.

Required background. You should be able to multiply small complex matrices, solve a two-by-two linear system, and work with complex conjugation. No abstract tensor notation is assumed. If you arrived from the mathematics diagnostic, return there after the final check and retry with different data.

Estimated study time. Roughly 1–2 hours for the examples, three exercises and re-check, assuming fluency with complex matrix multiplication, conjugation and two-by-two systems. This provisional estimate excludes prerequisite repair and further reading.

Vectors, covectors, and basis changes

Let $V$ be a finite-dimensional complex vector space with basis $\{e_a\}$. A vector $v\in V$ and a covector $\alpha\in V^*$ have expansions

$$v=v^a e_a, \qquad \alpha=\alpha_a e^a, \qquad \alpha(v)=\alpha_a v^a,$$

where $\{e^a\}$ is the dual basis, defined by $e^a(e_b)=\delta^a{}_b$. The positions of the indices record different types: $v^a$ are coordinates of a vector, while $\alpha_a$ are coordinates of a linear functional.

Choose a new basis

$$e'_a=e_b M^b{}_a$$

with $M$ invertible. Because the geometric objects do not change, their components must obey

$$v'^a=(M^{-1})^a{}_b v^b, \qquad \alpha'_a=\alpha_b M^b{}_a.$$

The opposite transformations make the pairing basis independent:

$$\alpha'_a v'^a =\alpha_b M^b{}_a(M^{-1})^a{}_c v^c =\alpha_b v^b.$$

This cancellation—not the visual appearance of one upper and one lower index—is what licenses the contraction. A systematic treatment of bases, duals, and maps appears in Axler 2015, chs. 1–3.

Linear maps and tensor type

A linear map $A:V\to W$ belongs to $W\otimes V^*$. In bases of $V$ and $W$, its components carry one index for each space:

$$w^i=A^i{}_a v^a.$$

The repeated $a$ index pairs $V^*$ with $V$; the free $i$ index says that the result lies in $W$. If the same space $V$ is used on both sides and its basis changes by $M$, the matrix changes by similarity,

$$A'=M^{-1}AM.$$

That formula is not a universal rule for every two-index array. A bilinear form $B:V\times V\to\mathbb C$ belongs to $V^*\otimes V^*$ and transforms instead as $B'=M^{\mathsf T}BM$. A Hermitian form transforms as $G'=M^\dagger GM$. The types decide the transformation law.

A quick index check prevents many errors:

1. Name the space or dual space associated with each index.
2. Pair only a space with its dual, unless an explicit form supplies a map between them.
3. Confirm that the free indices have the type of the claimed result.
4. Change basis once. A purported scalar should not change.

Forms, raising indices, and adjoints

A nondegenerate Hermitian form is conjugate-linear in its first argument:

$$\langle u,v\rangle_G=u^\dagger Gv.$$

For each $v$, the map $\alpha_v:w\mapsto\langle v,w\rangle_G$ is a linear covector. Its row of components is $v^\dagger G$, so

$$(\alpha_v)_a=\sum_b\overline{v^b}G_{ba}, \qquad \alpha_{cv}=\overline c\,\alpha_v.$$

Thus a nondegenerate Hermitian form identifies $V$ with its complex dual antilinearly. This is the ket-to-bra operation. It also has the correct covector transformation law:

$$(v')^\dagger G' =(M^{-1}v)^\dagger(M^\dagger GM) =v^\dagger GM.$$

The column $Gv$ instead represents the conjugate-linear functional $u\mapsto\langle u,v\rangle_G$; it is not the component column of an ordinary linear covector on a complex space. If $G$ is positive definite, it defines an inner product.

Spacetime index lowering uses a different structure: the real symmetric bilinear metric, or its bilinear extension to complex vectors. There,

$$v_\mu=g_{\mu\nu}v^\nu$$

is linear in $v$, with no complex conjugation. The Lorentzian metric is indefinite, so it does not define a Hilbert-space norm. Both identifications use additional structure; lowering an index is not typography. The treatment of forms and adjoints develops this distinction systematically.

The $G$-adjoint of $A$ is defined by

$$\langle u,Av\rangle_G =\langle A^{\dagger_G}u,v\rangle_G.$$

With the convention that the form is conjugate-linear in its first argument, this gives

$$A^{\dagger_G}=G^{-1}A^\dagger G.$$

Under the non-unitary basis change above,

$$G'=M^\dagger GM, \qquad A'=M^{-1}AM, \qquad (A')^{\dagger_{G'}}=M^{-1}A^{\dagger_G}M.$$

Thus self-adjointness is an invariant statement. For positive-definite $G$, it reduces to the matrix condition $A=A^\dagger$ in a $G$-orthonormal basis. For the role of adjoints and domains in quantum theory, see Hall 2013, ch. 2.

A QFT bridge: why a kinetic term is a scalar

For a scalar field $\phi$, the derivative $\partial_\mu\phi$ is a covector. The inverse metric supplies the map from covectors to vectors,

$$\partial^\mu\phi=g^{\mu\nu}\partial_\nu\phi.$$

The kinetic density

$$\frac12\partial_\mu\phi\,\partial^\mu\phi =\frac12g^{\mu\nu} (\partial_\mu\phi)(\partial_\nu\phi)$$

is a scalar because every tensor index is paired according to type. Without the metric, two covectors cannot be contracted canonically. Under a linear field redefinition $\phi^i=M^i{}_j\chi^j$, a multiplet kinetic matrix changes with the basis; the quadratic form does not:

$$\frac12\partial_\mu\phi^{\mathsf T}K\partial^\mu\phi =\frac12\partial_\mu\chi^{\mathsf T} (M^{\mathsf T}KM)\partial^\mu\chi.$$

This is the same invariant-content/component-array distinction used above. It later controls flavor mixing, gauge representations, propagator matrices, and the normalization of fields.

Exercises

1. Pairing under a non-orthogonal basis change

Let

$$M=\begin{pmatrix}1&1\\0&2\end{pmatrix}, \qquad v=\begin{pmatrix}2\\-1\end{pmatrix}, \qquad \alpha=\begin{pmatrix}3\\4\end{pmatrix},$$

where $v$ and $\alpha$ are the component columns of a vector and a covector in the original basis. Find $v'$ and $\alpha'$, then verify $\alpha'_a v'^a=\alpha_a v^a$.

Solution

Here

$$M^{-1}=\begin{pmatrix}1&-1/2\\0&1/2\end{pmatrix}.$$

The vector and covector laws give

$$v'=M^{-1}v =\begin{pmatrix}5/2\\-1/2\end{pmatrix}, \qquad \alpha'=M^{\mathsf T}\alpha =\begin{pmatrix}3\\11\end{pmatrix}.$$

The two pairings agree:

$$(\alpha')^{\mathsf T}v' =3\frac52+11\left(-\frac12\right)=2 =\alpha^{\mathsf T}v.$$

Using $M^{-1}$ on both objects would fail because a covector is not a second copy of a vector.

2. An adjoint in a non-orthonormal basis

Take

$$G=\begin{pmatrix}1&0\\0&2\end{pmatrix}, \qquad A=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$

Compute $A^{\dagger_G}$ and check the defining relation with $u=(1,i)^{\mathsf T}$ and $v=(2,1)^{\mathsf T}$.

Solution

Because $A^\dagger=A$,

$$A^{\dagger_G}=G^{-1}A^\dagger G =\begin{pmatrix}0&2\\1/2&0\end{pmatrix}.$$

Now $Av=(1,2)^{\mathsf T}$, so

$$\langle u,Av\rangle_G =u^\dagger GAv =(1,-i)\begin{pmatrix}1\\4\end{pmatrix} =1-4i.$$

Also $A^{\dagger_G}u=(2i,1/2)^{\mathsf T}$, and therefore

$$\langle A^{\dagger_G}u,v\rangle_G =(-2i,1/2)G\begin{pmatrix}2\\1\end{pmatrix} =1-4i.$$

The ordinary matrix $A$ is Hermitian but not self-adjoint with respect to this $G$; the form is part of the statement.

3. Diagnose an illegal contraction

Let $p_\mu\in V^*$ and $q_\mu\in V^*$. Explain what is missing from $p_\mu q_\mu$, and write a scalar after a nondegenerate metric is supplied.

Solution

Both displayed indices label covectors, so there is no natural pairing between them. Once an inverse metric $g^{\mu\nu}$ is given, the scalar is

$$g^{\mu\nu}p_\mu q_\nu=p_\mu q^\mu.$$

Writing two lower indices with the same label does not create the required map. In an orthonormal Euclidean basis the metric components happen to be a Kronecker delta, which can hide this structure.

Re-check and return

Choose a new invertible, non-unitary $2\times2$ matrix $M$, a positive Hermitian $G$, a map $A$, a vector $v$, and a covector $\alpha$. Without copying the formulas above, produce:

• the transformed components of all five objects;
• one invariant vector–covector pairing;
• one invariant quadratic form;
• the $G$-adjoint of $A$ before and after the basis change;
• the linear covector $\alpha_v=v^\dagger G$ associated with $v$, including its response to $v\mapsto iv$; and
• one sentence naming the type of every index in the calculation.

The repair is complete when the types determine the transformations and both scalar checks agree in the two bases. If the arithmetic works but the antilinearity of $v\mapsto\alpha_v$ remains implicit, review the forms section and repeat with a nondiagonal $G$ and a basis change with nonreal entries. If a purported scalar changes, return to the vector–covector pairing and add one object at a time.

Then retry the mathematics diagnostic with fresh matrices, or continue to Readiness. Core QFT uses this capability immediately in Classical fields and the action principle.

References

• Sheldon Axler, Linear Algebra Done Right, third edition, Springer, 2015, doi:10.1007/978-3-319-11080-6.
• Brian C. Hall, Quantum Theory for Mathematicians, Springer, 2013, doi:10.1007/978-1-4614-7116-5.
• Steven Roman, Advanced Linear Algebra, third edition, Springer, 2008, doi:10.1007/978-0-387-72831-5.