I have doubts in the statement that the partial or ordinary differentiation of tensor is not a tensor. The argument for this is that the partial differentiation of the tensor involves evaluating the transformation matrix at two neighbourhood points (say $P$ and $Q$) in the manifold and by the definition of tensor (the set of quantities that transform according to that rule where transformation matrix is evaluated at a point $P$) this is not the case. Thus, the partial differentiation of tensor is not tensor. Now my doubt is: what if the transformation matrices at those points are equal?

  • $\begingroup$ Maybe something more for the math-forum ? $\endgroup$
    – Nick
    Feb 8, 2014 at 10:16
  • $\begingroup$ On strictly logical grounds, "... is not necessarily a tensor" or even "... is almost never a tensor" is very different from "... cannot be a tensor". $\endgroup$
    – Mr. Nichan
    Sep 17, 2023 at 17:59

2 Answers 2


I'll write this as an answer so that the math is more clear.

So given an (p,q)-tensor $T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$, this one transforms as:

$$T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial x^{\nu'_1}}\cdots\frac{\partial x^{\nu_q}}{\partial x^{\nu'_q}}T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$$

Now if I were to take the partial derivative, then all of the factors in front of $T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ also contribute to extra terms since they don't need to be constant. This gives that I can't simply write:

$$\partial_{\rho'} T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial x^{\nu'_1}}\cdots\frac{\partial x^{\nu_q}}{\partial x^{\nu'_q}}\frac{\partial x^{\rho}}{\partial x^{\rho'}}\partial_{\rho}T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$$ and hence $T$ can't be a tensor!

That's why we define the covariant derivative:

$$\nabla_\mu T^\sigma{}_\rho=\partial_\mu T^\sigma{}_\rho+\Gamma_{\lambda\mu}^\sigma T^\lambda{}_\rho-\Gamma_{\mu\rho}^\lambda T^\sigma{}_\lambda$$ where we add connections for each index (plus-sign for upper indices, minus sign for lower indices). Now in order for the tensorial character of $\nabla_\mu$, the symbols $\Gamma_{\mu\nu}^\rho$ should be non-tensorial to cancel out the nontensorial character of the partial derivative (for the difference between the partial and covariant derivative see this post).

It can be shown that for the covariant derivatives to be a tensor, the transformation rule for the connections should be:

$${\Gamma~'}_{ij}^k=\frac{\partial x^p}{\partial y^i}\frac{\partial x^q}{\partial y^j}\Gamma_{pq}^r\frac{\partial y^k}{\partial x^r}+\frac{\partial y^k}{\partial x^m}\frac{\partial^2 x^m}{\partial y^i\partial y^j}$$ where the second term causes these symbols to be non-tensorial.

So to get back at your question, even if the transformation-matrices are equal, the non-tensorial character will show up due to the fact that we take derivatives of the elements of these matrices.

The ONLY case in which the partial differentiation is a tensor is in flat space! And every transformation with constant elements.

  • 3
    $\begingroup$ Hi @Nick: You can get bigger & centered equations by using $$ ... $$. Hope you don't mind my editing your post to center your equations. $\endgroup$
    – Kyle Kanos
    Feb 8, 2014 at 19:55
  • $\begingroup$ @KyleKanos, no problem at all. I appreciate the tip! $\endgroup$
    – Nick
    Feb 10, 2014 at 16:24
  • $\begingroup$ Additionally, the partial derivative of a scalar whether in flat or curved space, is of course, a real one-form. $\endgroup$
    – Ken Wang
    Aug 28, 2019 at 1:10

I) Nick's answer already correctly explains that the partial derivative $\partial_{\lambda}T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ of a $(p,q)$ tensor is in general not a tensor, in the sense that it does not transform covariantly under coordinate transformations.

II) In a coordinate-independent formulation, a $(p,q)$ tensor

$$\tag{1} T~\in~\Gamma (TM^{\times p}\times T^{*}M^{\times q}) $$

is a multi-linear map $T^{*}M^{\times p}\times TM^{\times q}\to \mathbb{R}$ where functions $f\in C^{\infty}(M)$ can be moved in and out of all its arguments, i.e.,

$$ f T(\eta_1, \ldots,\eta_p, X_1, \ldots,X_q)~=~T(f\eta_1, \ldots,\eta_p, X_1, \ldots,X_q)$$ $$\tag{2} ~=~\ldots~=~T(\eta_1, \ldots,\eta_p, X_1, \ldots,fX_q). $$

Here $\eta_1, \ldots,\eta_p \in \Gamma(T^{*}M)$ are one-forms/co-vector-fields, and $X_1, \ldots,X_p \in \Gamma(TM)$ are vector-fields.

III) It seems appropriate here to point out that exterior differentiation $\omega \mapsto \mathrm{d}\omega$ (which involves partial rather than covariant derivatives) does in fact take covariant tensors to covariant tensors, i.e. exterior forms to exterior forms. For a torsionfree tangentspace connection $\nabla$, this can be understood from the fact that partial and covariant differentiation of exterior forms yield the same result because the difference, i.e. the terms with Christoffel symbols, vanish under the antisymmetrization.

IV) Similarly, the Lie derivative $({\cal L}_{X}T)^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ of a tensor is again a tensor, as Nick points out in a comment below. Here $X\in\Gamma(TM)$ is a vector field. However note that the Lie derivative ${\cal L}_{X}T$ does not have the tensor-property (2) wrt. to the $X$-entry

$$\tag{3} {\cal L}_{fX}T ~\neq f{\cal L}_{X}T , \qquad f~\in~C^{\infty}(M).$$

  • $\begingroup$ +1 nice addition @Qmechanic! I believe that the same goes for the Lie-derivative ? Altough that one is a more complex construction. $\endgroup$
    – Nick
    Mar 9, 2014 at 21:25
  • 1
    $\begingroup$ @Nick: I updated the answer with your suggestion. $\endgroup$
    – Qmechanic
    Mar 9, 2014 at 22:26
  • $\begingroup$ Correction to the answer (v2): $X_p$ in the line below eq. (2) should read $X_q$. $\endgroup$
    – Qmechanic
    Mar 10, 2014 at 14:42

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