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Assume for example that you are given a (2,0) tensor $T^{\mu\nu}$ and you want to create a vector, i.e., a (1,0) tensor out of it. Is it possible to just fix an index of $T^{\mu\nu}$ while keeping the other free and thus define a vector? In other words, could you define a vector with components given by \begin{equation} X^\mu := T^{\mu 0}, \end{equation} with $\nu = 0$ fixed?

If I think about this at the level of multilinear maps, then it seems obvious to me. Let $V$ be an $n$-dimensional vector space with basis $(v_0,\ldots,v_{n-1})$ and denote the corresponding dual basis by $(v^0,\ldots,v^{n-1})$. The tensor $T$ is a multilinear map $T\colon V^* \times V^* \to \mathbb{R}$. Define the map $X\colon V^* \to \mathbb{R}$ as \begin{equation} X(w):= T(w,v^0). \end{equation} Then obviously $X$ inherits linearity in its argument from the linearity in the first argument of $T$. The components of $X$ are just as desired \begin{equation} X^\mu:= X(v^\mu) = T(v^\mu,v^0) =: T^{\mu 0}. \end{equation} And upon change of basis we get the correct transformation law \begin{equation} X'^\mu = X(v'^\mu) = X(\lambda{^\mu}{_\nu} v^\nu) = \lambda{^\mu}{_\nu} X(v^\nu) = \lambda{^\mu}{_\nu} X^\nu = \lambda{^\mu}{_\nu} T^{\nu 0}, \end{equation} where really only the first index of $T$ is transformed.

If I however blindly insert \begin{equation} X'^\mu = T'^{\mu 0} \end{equation} then it looks like I would have to transform the second index of $T$ as well with the transformation law for (2,0) tensors. It seems to me that this is just a problem with notation however.

Coming to a more specific problem, let us consider a smooth manifold structure. Is it now valid to define the vector \begin{equation} X^\mu(x) := \phi(x)\delta{^\mu}{_0}, \end{equation} where $\delta{^\mu}{_\nu}$ is the Kronecker-delta, a (1,1) tensor, and $\phi(x)$ some arbitrary scalar depending on the chart coordinates $x$? I.e., is this really a 4-vector that transforms correctly?

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    $\begingroup$ You and I accept $$ T'^{\,\mu\,\rho}={\lambda^\mu}_\nu {\lambda^\rho}_{\sigma}T^{\,\nu\,\sigma}\,. $$ Therefore I do not think that it all works because $$ X^\nu:=T^{\,\nu\,0}\,,\quad X'^{\,\mu}=T'^{\,\mu\,0}={\lambda^\mu}_\nu {\lambda^0}_{\sigma}T^{\,\nu\,\sigma}\not={\lambda^\mu}_\nu\, T^{\,\nu\,0}\,. $$ $\endgroup$
    – Kurt G.
    Commented Feb 26 at 12:33
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    $\begingroup$ I don't have the time to write a full answer just now, but what you've discovered is that there's a difference between the definition of a tensor as "a multilinear map from $V$ & $V^*$ to the real numbers" and "a set of components that transforms according to some set of coordinate transformation rules." For another example of this, read the brief discussion in Wald where he discusses why the Christoffel symbols can be thought of as a tensor even though they don't obey the coordinate transformation rules. $\endgroup$
    – Michael Seifert
    Commented Feb 26 at 12:52
  • $\begingroup$ Thank you both for the comments. @MichaelSeifert I will have a look at Walds book. I'd be happy if you found the time later to elaborate your answer. $\endgroup$
    – Burgulence
    Commented Feb 26 at 14:18
  • $\begingroup$ It's a busy week for me but I will try to carve out some time for it. FYI the Wald paragraph I was thinking of is immediately following Eq. (3.1.15), where he defines the Christoffel symbols. $\endgroup$
    – Michael Seifert
    Commented Feb 26 at 15:14

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Yes you can but only once. In one chosen coordinate system and at on point in time.

If after that you want to transform things to a new coordinate system, or if there is a transformation because a physical operation is performed, then you must transform the vector that you created like a vector, whereas the coordinates $T^{\mu0}$ transform differently (they even need the rest of the tensor $T$ in order to know how to transform).

So you can create a vector $X^\mu=T^{\mu0}$ but it will usually not remain equal to $T^{\mu0}$ when affected by the usual things that we do in physics calculations. But if you are careful about that it may still serve some purpose.

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  • $\begingroup$ So is it like: we define the components for the vector in one coordinate system as such and such value and then require it to transform like a vector for a change of coordinates? So it would be a vector by requirement. $\endgroup$
    – Burgulence
    Commented Feb 26 at 14:16
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    $\begingroup$ As opposed to being a vector "by inspection", perhaps. It does not by itself transform as a vector. However, if we create it like:$$ X^\mu = T^{\mu\nu} Y_\nu$$ where you just make the choice (1,0,0,0) for $Y$, you can also say that if you transform $Y$ as a co-vector and also transform $T$ as a tensor, then the defining relation remains valid after transformation. But it remains $X^\mu=T^{\mu\nu} Y_\nu$, it does not remain $X^\mu=T^{\mu0}$ $\endgroup$
    – Jos Bergervoet
    Commented Feb 26 at 15:01

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