Working through Schutz "First course in general relativity" + Carroll, Hartle and Collier, with some help from Wikipedia and older posts on this forum. I am confused about the gradient one-form and whether or not it is normal to a surface. In the words of Wikipedia (gradient): If f is differentiable, then the dot product (∇f )x ⋅ v of the gradient at a point x with a vector v gives the directional derivative of f at x in the direction v. It follows that in this case the gradient of f is orthogonal to the level sets of f. The first sentence is almost clear. What do they mean by "dot product"? Is this a scalar product, which I think requires a metric, or a contraction, which does not? When written out, the formula for the directional derivative contains no metric and is clearly the contraction of a vector and the gradient. It's the second sentence that gets me. Just how it "follows" that the gradient is orthogonal to the level sets does not jump out at me. (Sorry, I don't know how to write formulas on the forum.) In good ol'-fashioned 2d Euclidan geometry, if y = f(x), then the gradient dy/dx is the slope of the tangent to the curve at a particular value of x. So how do we get from tangents to orthogonals? To continue, Schutz (64) talks about the model of one-forms as surfaces crossed by vectors. He says: "... the one-form's surfaces are straight and parallel.This is because we deal with one-forms at a point, not over an extended region: 'tangent' one-forms in the same sense as tangent vectors." So now one-forms are tangents, even tho the gradient is normal. As Schutz (66) says 2 pages later: "A one-form is said to be normal to a surface if its value is zero on every vector tangent to the surface." By "on", I presume he means "associated with". That, at least, seems clear. A normal to a surface has no components on a tangent to the surface. Schutz again (72): "A vector is said to be normal to a surface if its associated one-form is a normal one-form." I find all this very confusing and difficult to visualize. Yet visualize I must.