# Is a 1D vector also a scalar?

A vector in one dimension has only one component. Can we consider it as a scalar at the same time?

Why time is not a vector, although it can be negative and positive (when solving for time the kinematics equation for example)?

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A scalar is defined to be invariant under transformations of the coordinate system. Thus, a vector in one dimension is not a scalar.

Time is a "parameter", or a component of a 4-vector in special relativity. In classical mechanics, it is essentially a one-dimensional vector.

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[I realize this in an old question, but this comment is meant for future readers.] Your first line is a bit vague. You're talking about orthogonal transformations, living in $O(1) = \{ \pm 1 \}$. Restricting as usual to $SO(1)$ leaves you with nothing but the identity transformation, so any 1D vector would indeed be a scalar. –  Vibert Feb 27 '13 at 22:28

A scalar with a unit is a 1-dimensional (axial) vector; changing the basis corresponds to changing the unit.

A number (without a unit) is not a 1-dimensional vector in the terminology used by physicists. However, it is a 1-dimensional vector in the terminology used in linear algebra.

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A vector in a $1D$ space is not a scalar. But if we choose a basis (which in this case consists only in one vector, say $E$), any other vector is of the form $vE$, with $v$ a scalar. So we can identify the $1D$ vector space with $\mathbb R$, but the identification depends on the choice of $E$.

In the case of the time, things are similar. For the Minkowski spacetime, consider an orthonormal basis, consisting in three spacelike vectors and one timelike vector. This basis allows us to express any event in terms of three space coordinates, and one time coordinate. The time coordinate is a scalar, and can indeed be positive or negative. For curved spacetime, the things get more difficult. The spacetime is no longer a vector space. We use coordinates, which are no longer obtained from vector frames. Sometimes they can't even be global, so we have to take them local (limited to a subset of the spacetime).

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The perspectives of the other answers I do not think are the correct ones. Here is one from the perspective of representation theory.

Whether a quantity is a "scalar" or a "vector" (or something more exotic) is a question of what representation of the group of isometries it resides in. For n-dimensional Euclidean space, this is the group O(n). For n=1, O(n) has just the elements 1 and -1. A vector acts nontrivially under -1, while a scalar is unchanged.

Speaking more broadly, we can consider antisymmetric tensor fields (sections of the exterior powers of the tangent bundle). The top exterior power, the so-called tangent frames (or if you prefer their duals, the volume forms), are in bijection with the group of scalars if our space is orientable. That is, fixing an orientation (which is a global section of this bundle) O, every other top rank tensor is of the form f(x)O for some scalar function f. If we're in Euclidean space, only the parity transformation -1 can act nontrivially on one of these. It acts trivially iff the dimension is even, so scalars are top tensor fields in even dimensions and psuedoscalars are top tensor fields in odd dimensions.

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Your view is in conflict with the standard convention that allows one to talk about polar and axial vectors, which have different behavior under eflection. A scalar is an axial vector in 1D. –  Arnold Neumaier Sep 5 '12 at 8:53
In my answer, polar and axial vectors correspond to 1-frames (ordinary vectors) and n-1-frames. –  Ryan Thorngren Sep 5 '12 at 19:30