In relativity, given a metric tensor $g_{\mu\nu}$ and its inverse $g^{\mu\nu}$, by definition of the inverse, we have the relation $g_{\mu\rho}g^{\rho\nu} = \delta_\mu{}^\nu$. In matrix form, $\delta_\mu{}^\nu$ can be written as

$$ \delta = (\delta_\mu{}^\nu) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, $$

which, if we forget about the positioning of the indices for a while (can you say "index signature"?), is a symmetric matrix. Since this holds for any metric, it feels like this indicates that $\delta_\mu{}^\nu$ is a symmetric tensor.

That $\delta$ is symmetric can be expressed as $\delta^\text{T} = \delta$. But how do we express that $\delta_\mu{}^\nu$ is symmetric? Neither $\delta_\mu{}^\nu = \delta_\nu{}^\mu$ nor $\delta_\mu{}^\nu = \delta^\nu{}_\mu$ makes sense to me. Does it make sense to say that $\delta_\mu{}^\nu$ is symmetric, or can you really only speak about symmetry of a tensor when it has two indices which are either both covariant or both contravariant (can you say that the indices have the same "variance"?)?

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    $\begingroup$ Note that the transpose of an arbitrary linear map $A: V \to V$ is not actually a map from $V \to V$; it's a map from $A^T: V^* \to V^*$, where $V^*$ is the dual space to $V$. So saying that $A = A^T$ is something of a category error. That said, my mind really wants there to be some sense in which $\delta = \delta^T$, so I'll be watching this question with interest. It might have something to do with the adjoint instead, but that depends on the existence of an inner product on $V$ and I'm a bit surprised that we can't define a notion of symmetry without it. $\endgroup$ Mar 25, 2021 at 13:59
  • $\begingroup$ Also, if you don't get an answer after a few days, you might consider re-posting this on Mathematics instead. $\endgroup$ Mar 25, 2021 at 14:00
  • $\begingroup$ I don't really see what is wrong with $\delta_{\mu}^{\; \nu}=\delta_{\nu}^{\; \mu}$, could you elaborate? $\endgroup$
    – jacob1729
    Mar 25, 2021 at 14:00
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    $\begingroup$ @jacob1729 In the left hand side, $\mu$ is a covariant index and $\nu$ is a contravariant index, which doesn't match their qualities in the right hand side. I'm not completely sure this makes the equation nonsensical, but I can't remember having seen any equation before where the qualities of the indices don't match up between the LHS and RHS. $\endgroup$ Mar 25, 2021 at 14:25
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    $\begingroup$ The "Kronecker symbol" is a constant tensor of rank (1,1) on a differentiable manifold, as of such its components in a coordinate basis can be simply denoted by $\delta_{\mu}^{\nu}$, so the exact index slot placement is not important (the unit matrix is symmetric), i.e. $\delta_{\mu}^{\nu} \equiv \delta_{~~\mu}^{\nu} \equiv \delta_{\mu}^{~~\nu}$. $\endgroup$
    – DanielC
    Mar 25, 2021 at 16:33

2 Answers 2


There can be different notions of "symmetric". Also, keep in mind that $\delta$ is just the index form of the identity operator.

The first one pertains to the transpose of an operator, which is just the adjoint for a real case. Generally, for the transpose of a tensor, we have $$\left<v,M w\right>=\left<M^T v,w\right>\,,$$ or in index notation $$\left<v,M w\right>=v^\mu g_{\mu\nu}\left(Mw\right)^\nu=v^\mu g_{\mu\nu} M^\nu_{\phantom{\nu}\rho} w^\rho\\ =\left<M^T v,w\right>=\left(M^T v \right)^\lambda g_{\lambda\rho} w^\rho= \left(M^T\right)^\lambda_{\phantom{\lambda}\mu}v^\mu g_{\lambda\rho} w^\rho\,.$$ Since this is true for all $v$ and $w$, we can deduce that $$\left(M^T\right)^\kappa_{\phantom{\lambda}\mu}=\left(M^T\right)^\lambda_{\phantom{\lambda}\mu}g_{\lambda\rho}\cdot g^{\kappa\rho}=g_{\mu\nu} M^\nu_{\phantom{\nu}\rho} \cdot g^{\kappa\rho}=M_\mu^{\phantom{\mu}\kappa}\,.$$ This is the first of your guesses, and it does make sense. Thus, in this sense, $\delta$ is indeed "symmetric", i.e. equal to its transpose -- note that this is somewhat trivial, because it means that the identity operator is selfadjoint.

On the other hand, in the context of GR or group representations, "symmetric" is ususally understood as relating to a pair (or more) of same-type indices, such as $M_{\mu\nu}=M_{\nu\mu}$. This does not apply to your case.)

(UPDATE: Note that here I have used a slightly different terminology than J.Murray in his answer: I consider the transpose to be the adjoint in a real vector space with a bilinear scalar product (as opposed to a complex space wth sesquilinear product). In my experience, this seems to be standard, and then it is not wrong to ask whether $A^T=A$ -- indeed, orthogonal transformations are generally those for which $O^T=O^{-1}$, i.e. where $O^T$ acts on $V$; this is of course the real analogue of unitary transformations, $U^\dagger=U^{-1}$ in complex spaces.

Note also that without a scalar product, you cannot identify endomorphisms of $V$ and those of $V^*$ componentwise. A choice of dual basis is not enough to provide a canonical (i.e. basis-indepedent) isomorphism between $V$ and $V^*$.)


The quick answer is that yes, it does make sense, if we interpret symmetric to mean self-adjoint. Symmetry of a bilinear form $g$ (or higher rank tensor) means $g(X,Y)=g(Y,X)$, but symmetry of a linear operator $A$ means $\langle X,A(Y)\rangle = \langle A(X),Y\rangle $ where the brackets denote some chosen inner product. In that sense, symmetry of a linear operator requires an inner product structure.

When we say that an operator is symmetric when it equals its transpose, we are (as pointed out by Michael Seifert's comment) making a category error. However, this can be understood as saying that $A$ and $A^\mathrm T$ are equal component-wise. While this does not strictly require an inner product, it is equivalent to choosing one and then expressing the components of $A$ and $A^\mathrm T$ in an orthonormal basis.

Before I climb up onto my soapbox for the long answer, it's important to make the point that I'll be talking about linear maps - functions between vector spaces, with a given domain and a given codomain. I distinguish this from a matrix, which is just a rectangular array of numbers which provides a convenient way to organize the components of a linear map in some given basis.

The long answer goes as follows:


Let $V$ and $W$ be vector spaces and $V^*$ and $W^*$ their algebraic duals - that is, the set of linear maps from $V$ and $W$ to $\mathbb R$ (the extension to $\mathbb C$ is straightforward).

Given a linear operator $A:V\rightarrow W$, we define its transpose $A^\mathrm T:W^* \rightarrow V^*$ as follows. For every $\omega\in W^*$ and $v\in V$,

$$(A^\mathrm T \omega)(v) := \omega\big(Av)$$

In component notation, $$(A^\mathrm{T})_i^{\ \ j} \omega_j v^i = \omega_j A^j_{\ \ i} v^i \iff (A^\mathrm T)_i^{\ \ j} = A^j_{\ \ i}$$

If we arrange the components of $A$ and $A^\mathrm T$ in rectangular grids and let the first index be the "row" index and the second be the "column" index, then the grid form of $A^\mathrm T$ is just the grid form of $A$ turned on its side.

Note that $V$ and $W$ could be completely different spaces of different dimensions; in array form, $A$ is generically an $n\times m$ matrix and $A^\mathrm T$ is then an $m\times n$ matrix.


Until now, $V$ and $W$ have been bare vector spaces with no additional structure. Now we specialize to the case where they are both Hilbert spaces, equipped with inner products $\langle \cdot,\cdot\rangle_V$ and $\langle \cdot,\cdot \rangle_W$ respectively. We also assume that they are finite-dimensional, as the infinite-dimensional case comes with substantial technical subtleties. Given a linear operator $A:V\rightarrow W$, we define its adjoint $A^\dagger: W\rightarrow V$ as follows. For every $v\in V$ and $w\in W$,

$$\langle A^\dagger w, v\rangle_V = \langle w, A v\rangle_W$$ In component form, if the inner products are given by metric tensors $(g_V)_{ij}$ and $(g_\mathrm W)_{ij}$, we have $$(g_V)_{i\ell}(A^\dagger)^i_{\ \ j}w^j v^\ell = (g_W)_{ji} w^j A^i_\ell v^\ell \iff (A^\dagger)^i_{\ \ j} = (g_V)_{j\ell} A^\ell_{\ \ m}(g_W)^{mi}$$

or in matrix multiplication form, $$(A^\dagger)^i_{\ \ j} = \left(g_V \cdot A \cdot g_W^{-1}\right)_j^{\ \ i}$$

Relationship between $A^\mathrm T$ and $A^\dagger$

To summarize so far, the transpose of a linear map always exists, requiring no extra structure to define, while the adjoint requires Hilbert space structures.

When the adjoint exists, it is related to the transpose in the following way. For the moment, I restrict my attention to linear operators $A:V\rightarrow V$ on a single Hilbert space; the generalization $A:V\rightarrow W$ is conceptually straightforward.

The inner product defines a canonical isomorphism between $V$ and $V^*$; to each vector $v\in V$ we associate a dual vector $\tilde v \in V^*$ where $$\tilde v := \langle v,\bullet\rangle$$

In component form, $\tilde v_i = g_{ij} v^j$. Given this bijection, we can now state the relationship between the transpose and adjoint:

$$A^\mathrm T \tilde v = \langle A^\dagger v,\bullet\rangle$$

In component form, $$(A^\mathrm T)_i^{\ \ j} = g_{i\ell} (A^\dagger)^\ell_{\ \ k} g^{kj}$$ and in matrix form, $$A^\mathrm T = g \cdot A^\dagger \cdot g^{-1}$$

In other words, the action of the transpose on a dual vector can be understood as first raising the index on the covector, then acting on it with $A^\dagger$, and finally lowering the index again.

As an important note, if we choose an orthonormal basis such that $g_{ij}=\delta_{ij}$, then the components of the transpose are numerically equal to the components of the adjoint. For example, $(A^\mathrm T)_1^{\ \ 2} = (A^\dagger)^1_{\ \ 2}$. If we use the row-column convention for writing out these components in square arrays, then the matrix representations of these operators are identical - but it's important to remember that (i) this assumes that the basis in question is orthonormal, and (ii) we should not confuse the operators with their components.

What is Symmetry?

There are several definitions of symmetry which are usually quite closely related.

  1. A linear operator on a (finite-dimensional) vector space equipped with an inner product is called symmetric if $A= A^\dagger$. In physics, we often call such operators "hermitian," especially in quantum mechanics.

  2. On the other hand, in elementary linear algebra we usually define a linear operator $A:V\rightarrow V$ as being symmetric if $A=A^\mathrm T$. However, note that based on the formalism developed above, that doesn't make sense; $A$ and $A^\mathrm T$ act on different spaces. However, since $V$ and $V^*$ are isomorphic and a choice of basis $\{\hat e_i\}$ for $V$ induces a choice of basis $\hat \{\epsilon^i\}$ on $V^*$ such that $\hat \epsilon^i(\hat e_j)= \delta^i_j$, we may define $A$ as symmetric if $A$ and $A^\mathrm T$ are equal component-wise. That is, they are not the same as maps, but they have the same components as long as we choose canonically-related bases $\{\hat e_i\}$ and $\{\hat \epsilon^i\}$ for the two spaces.

  3. Finally, a bilinear form $g$ is symmetric if, for all vectors $v,w\in V$, we have that $g(v,w)=g(w,v)$. In component form, this is just the statement that $g_{ij}=g_{ji}$.

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    $\begingroup$ Excellent answer! I think you have a typoe in your last point #2, though; do you mean "$A$ and $A^T$ act on different spaces?" $\endgroup$ Mar 25, 2021 at 21:56
  • $\begingroup$ @MichaelSeifert I knew one of those would slip through :) Thanks for the catch (and the kind comment). $\endgroup$
    – J. Murray
    Mar 25, 2021 at 22:04
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    $\begingroup$ Hmm, the wiki article en.wikipedia.org/wiki/Hermitian_adjoint says that the Hermitian adjoint is only defined for linear operators on a single Hilbert space, but the section en.wikipedia.org/wiki/… says that the concept works for general linear maps between (possibly different) Hilbert spaces. Is there any reason why you can't use the two different Hilbert spaces' inner products for the adjoint (and so two different $g$ matrices, potentially of different sizes, in your matrix version)? $\endgroup$
    – tparker
    Mar 26, 2021 at 1:39
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    $\begingroup$ @tparker That's a nice catch. No, there's no reason why you can't make that generalization. As the converse of what I wrote above, we can understand the adjoint as the map obtained by lowering the index, applying the transpose, and then raising the index again; since the transpose doesn't require the spaces to be the same, the adjoint in principle doesn't either. $\endgroup$
    – J. Murray
    Mar 26, 2021 at 2:39
  • $\begingroup$ Thanks. You may want to also edit out your requirement of operators later in your answer. $\endgroup$
    – tparker
    Mar 26, 2021 at 3:35

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