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As far as I understand it, the notion of direction is not an intrinsic property of a vector in an abstract vector space; if the space is equipped with an inner product then one can determine angles between vectors and hence then gain a notion of a vector have direction (as well as a magnitude, as an inner product implies that the vector norm exists). Have I understood this correctly?

If so, when defining tangent vectors on a manifold, is the motivation for considering them as equivalence classes of curves passing through a given point (with equal derivatives, to first-order, in some coordinate chart), that the directional derivatives that arise from this approach satisfy the vector space axioms and thus are sufficient to unambiguously define vectors at each point on the manifold?

My confusion stems from the fact that if one chooses a curve passing through a particular point on a manifold and then takes the directional derivative on some differentiable function $f$ defined on $M$, then aren't we implying through this that defined via such curves intrinsically have a notion of direction at each point on the manifold?

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Correct, for example in $\mathbb{R}^3$ our intuitive idea of direction is with respect to the standard basis.

Forgetting about general manifolds, consider the directional derivative of a function $f$ defined on a 3-dimensional vector space (at a point along a curve). It is completely independent of what basis of the vector space you might like. You are concerned with how $f$ changes relative to the curve.

Actually tangent vectors can be defined on abstract manifolds that have no inner product - tangent spaces necessarily have nothing to do with direction.

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  • $\begingroup$ Ok, so is the motivation behind defining vectors in terms of curves that in doing so we determine objects that satisfy the vector space axioms (and direction doesn't come into it)? Another couple of questions, obviously every vector space has at least one basis that one can expand any vector in the space onto, but this itself would not define a notion of direction unless one defines an isomorphism between the vector space and $\mathbb{R}^{n}$?! $\endgroup$
    – Perpetual learner
    Mar 5, 2015 at 14:51
  • $\begingroup$ Also, is it correct to say that in order to determine how a function changes relative to a given curve we must evaluate the function along that curve, i.e. given a curve $\gamma$ paramterised by $t$ we can determine how a function $f$ changes relative to this curve by composing the function with the curve, $f\circ\gamma$ such that for each value of $t$ we determine a value of the function at the point $\gamma (t)$ along the curve, $(f\circ\gamma) (t)=f(\gamma (t))$. We can then determine the rate of change of $f$ relative to $\gamma$ by taking the derivative $\frac{d}{dt}(f\circ\gamma (t))$. $\endgroup$
    – Perpetual learner
    Mar 5, 2015 at 14:55
  • $\begingroup$ Say you have a surface in $\mathbb{R}^3$. If you pick a point, you probably know what we mean by tangent plane. It's often the plane that locally only touches the surface at one point. The directional derivative definition is just one way of defining this precisely. You take equivalence classes because different functions can have the same behavior near the point, they have the same derivative. For an abstract vector space, as you said, you need an inner product for direction. My point with R^3 was "direction" is always relative to a basis. $\endgroup$
    – mathematician
    Mar 5, 2015 at 15:15
  • $\begingroup$ 2nd comment: You would want "How f changes relative to $\gamma$" to not depend on the parametrization of $\gamma$. You can use an arc length parametrization for the quantity to be intrinsic to the curve. $\endgroup$
    – mathematician
    Mar 5, 2015 at 15:17
  • $\begingroup$ How does one show that it's independent of the parametrisation? $\endgroup$
    – Perpetual learner
    Mar 5, 2015 at 15:31

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