When studying tensor representations of the special unitary group $\mathrm{SU}(N)$, one usually introduce two types of vectors $\psi^i, \psi_i$ (see this website or Chapter IV.4 in Zee's book Group Theory in a Nutshell for Physicists). The one with upper index $\psi^i$ is ordinary, which transforms under $U \in \mathrm{SU}(N)$ as

$$ \psi^i \to {U^i}_j \psi^j $$

while the one with lower index $\psi_i$ is defined to be the complex conjugate of $\psi^i$

$$ \psi_i := \bar{\psi}^i $$

More confusions arise if I dig deeper. For example, I cannot decide whether the transpose of ${U^i}_j$ should be denoted as ${(U^T)_j}^i$ or ${(U^T)^j}_i$, and I am not clear on the relation between ${U^i}_j$ and ${U_i}^j$.


  • What is the full machinery behind this lower/upper index notation? And why raising/lowering the index is accompanied by a complex conjugation?

  • Is there a generalization to deal with the indefinite unitary groups $\mathrm{SU}(p,q)$ (with $p+q = N$)?

  • $\begingroup$ $U^\dagger$ is transpose and complex conjugate. The condition to be unitary is $(U^\dagger)^i_{\;j}U^j_{\;k}=\delta^i_{\,k}$. If you want to write the transformation of the covariant objects correctly, you could write $a_i(U^\dagger)^i_{\;j}$. Then for transformed coordinates you get $(a')_i(b')^i=a_ib^i=\langle a^*,b\rangle$. $\endgroup$ Oct 14, 2023 at 17:08
  • $\begingroup$ Since you are reading a physicist text, it is the same thing as the bra-ket formalism: In the defining (left) representation on $\mathbb{C}^n$ $U$ acts by $U\lvert\psi\rangle$ on the ket's (a single upper index = a column vector = a ket) and the correspondingly $\langle\psi\rvert U$ (a single lower index = a row vector = a bra). If you want to switch that back to left action you of course do $U^{-1}\cdot\langle\psi\rvert$ which is why there is the transpose. The complex conjugation comes from the inner product. $\endgroup$ Oct 14, 2023 at 22:39
  • $\begingroup$ @user10354138 Do you mean that the lower index vectors correspond to the dual representation of SU($N$)? $\endgroup$ Oct 15, 2023 at 10:30
  • $\begingroup$ Yes. Take a column vector space and a compatible matrix group, both with complex entries. The group can act on the vectors via the standard matrix-vector product in only two ways faithfully, i.e., as representation, so that $\rho(UV)\psi=\rho(U)\rho(V)\psi$. Either directly via $\rho(U)=U$ or per the complex conjugate, $\rho(U)=\bar U$. For unitary matrices one can write this complicated as $\bar U=U^{-T}$, making the dual also contragredient, "against the gradient" (remember that the gradient is defined via the scalar product, so in the Hermitean case involves complex conjugation). $\endgroup$ Oct 15, 2023 at 10:45
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    $\begingroup$ Yes, for $SU(n), n>2$ its defining representation is conjugate to its dual but not isomorphic. Whereas, for $SO(n)$ its defining representation is conjugate to itself (indeed it is real) and also dual to itself. $\endgroup$
    – Callum
    Oct 15, 2023 at 17:24

1 Answer 1


Remember that usually in the tensor or Ricci calculus, upper indices stand for covariant components, like the coordinates of a vector or the coordinate functionals in a dual basis. And lower indices stand for covariant components such as the vectors in a basis or the coefficients of a linear functional.

With complex or Hermitean vector spaces, an additional notational difficulty arises in the slight difference between the trace and the scalar product, in that the trace is used as bilinear, while the scalar product is anti-linear in one argument, the first if compatible with bra-ket notation.

So if you define a linear functional from a vector via duality using the scalar product as $$\alpha(v)=\langle a\mid v\rangle,$$ the coefficients of the linear form are $$\alpha_i=(a^i)^*.$$

Note that the asterisk has some ambiguous interpretation. In $U^\dagger$ or more seldom $U^H$ it is without doubt that both transposition and complex conjugation are applied. In $\overline U$ it is also clear that the entries get conjugated without changing their position. In mathematics $U^*$ denotes the adjoint operator, in the Hermitean case equal to $U^\dagger$. In other contexts $z^*$ is readily used as simple complex conjugate $\bar z$, so one would have to check in the immediate context. So it could be that $(a^i)^*$ is also quite naturally $(a^*)_i$.

$U^T$ or $U^H$ are still matrices or linear operators on the same vector space, so the index positions remain the same as in $U\in {\rm Hom}(V)\simeq V\otimes V^*$. In general one considers all tensor products of a space with itself or its dual space and of the same signature as isomorphic, so that one does not need to coordinate the index positions between upper and lower indices, inserting spaces etc.

I do not see how any of this would change for indefinite scalar products, except that you would have to include the matrix of the scalar product, $⟨u∣v⟩=\bar u^i\eta_{ij}v^j$, so that now $(a^*)_j=\overline{( a^i)}\eta_{ij}$. This means that the adjoint construction already contains the (sign) modifications from the scalar product.

  • $\begingroup$ What are the upper indices co-varying with? In the ordinary case, the covariance is with respect to a change of basis; but from this perspective no complex conjugation is involved. The conjugation introduced is more than to merely make the expression of the hermitian form look fancier. $\endgroup$ Oct 16, 2023 at 0:43
  • $\begingroup$ They are co-varying with the complex conjugate of the usual basis vectors. This mainly serves to be able to include constructions with the hermitean scalar product into the tensor calculus that is mainly multi-linear in its operations. $\endgroup$ Oct 16, 2023 at 9:20
  • $\begingroup$ They are co-varying with the complex conjugate of the usual basis vectors. This mainly serves to be able to include constructions with the hermitean scalar product into the tensor calculus that is mainly multi-linear in its operations. // It seems like a record on repeat. Most of the Index or Ricci calculus is a theory or system of notation that evolved from quantum and gravitiational calculations in the 1920-30ies. The new theory drove the need for suitable notation, and properly adapted notation highlights parts of the theory that are "natural" vs. those that are more questionable. $\endgroup$ Oct 16, 2023 at 10:43
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    $\begingroup$ That was to say, I'm sorry for the repeats, there just is no more inherent content in this. You get elementary matrix representations of (unitary) groups that are left representations, matrix times column vector, that are contravariant and contragredient. And right representations, row vector times matrix, that are covariant and cogredient. // Now dual vector spaces, matrices and tensor products are still vector spaces, thus isomorphic to some column vector spaces. Changing the format of the representation to suit this is the source of much confusion. $\endgroup$ Oct 16, 2023 at 11:00

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