Why are dimensions always at right angles?

1,228
188
Don't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...
 

WWGD

Science Advisor
Gold Member
4,174
1,742
Don't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...
Riemannian metrics of the form ##adx^2+bdy^2+cdz^2 ## where ##a,b,c## not all ##1## refer to, or depict non-Euclidean geometries, or geometries where

##a^2+b^2=c^2## does not quite hold. But, yes, in the Calculus done in Euclidean space, the generalized Pythagorean theorem does hold. But the Pythagorean theorem does not hold in general when your Calculus is done on a general manifold.
 

Mark Harder

Gold Member
231
52
Riemannian metrics of the form ##adx^2+bdy^2+cdz^2 ## where ##a,b,c## not all ##1## refer to, or depict non-Euclidean geometries, or geometries where

##a^2+b^2=c^2## does not quite hold. But, yes, in the Calculus done in Euclidean space, the generalized Pythagorean theorem does hold. But the Pythagorean theorem does not hold in general when your Calculus is done on a general manifold.
That looks like a generalized inner product using a quadratic form:
d×(x,y,z)⋅((a,0,0),(0,b,0),(0,0,c))⋅(x,y,z)T, where (...) indicates a row matrix and ((...),(...)) is a matrix, in this case a diagonal matrix. Is this the right way to think of the metrics you refer to?
 

WWGD

Science Advisor
Gold Member
4,174
1,742
That looks like a generalized inner product using a quadratic form:
d×(x,y,z)⋅((a,0,0),(0,b,0),(0,0,c))⋅(x,y,z)T, where (...) indicates a row matrix and ((...),(...)) is a matrix, in this case a diagonal matrix. Is this the right way to think of the metrics you refer to?
Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix ##g(X_i, X_j)## where the {## X_i ##} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.
 

Mark Harder

Gold Member
231
52
Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix ##g(X_i, X_j)## where the {## X_i ##} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.
I occurred to me after I wrote that post that by dx2, you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...
 

WWGD

Science Advisor
Gold Member
4,174
1,742
I occurred to me after I wrote that post that by dx2, you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...
No problem, feel free to ask, I will do my best, and ask for clarification if needed.
Formally, ## dx^2 ## is ## dx\otimes dx ##, the tensor product of dx with itself. And dx,dy,dz are usually define as covectors, as a basis dual to the basis ## \partial/\partial x, \partial / \partial y, \partial /\partial z ## of basis tangent vector fields meaning ##dx( \partial /\partial x) =1 , dx (\partial /\partial y)=0 , i.e., dx_i (\partial / \partial_j):= \delta^i_j## , etc. This means that linear operators on tangent vectors are written as ## a(x,y,z)dx+b(x,y,z)dy+c(x,y,z)dz ## (notice that in the expression for your metric , you can have "mixed terms" ##dx \otimes dy ## , etc., but this is a somewhat-simplified version) , i.e., as linear combinations of these covector basis elements. dx, dy, dz are differential forms, i.e., linear maps defined on tangent vectors. This allows you to compute the length of curves.
Maybe more precisely, this metric or second fundamental form, and it allows you to find the length
of a parametrized curve under a given choice of metric. Here ## dx \otimes dx (v_1,v_2):=dx(v_1)dx(v_2) ## , meaning scalar multiplication, for vectors ## v_1, v_2 ##. This allows you to define forms on n-ples of vectors ( in this case, on pairs of vectors ), i.e., given linear maps ##dx, dy## defined on vectors ##v_1, v_2## , you can form a bilinear map ## dx \otimes dy ## defined on the ordered pair of vectors ## (v_1, v_2) ##, and so on, i.e., you can define k-linear maps on ordered k-ples of vectors. The collection of linear, bilinear,..k-linear maps defined on the exterior product of a vector space is called the exterior algebra of the exterior product. This can be defined on (the exterior product of) any vector space, not just the tangent space , exterior powers of the tangent space.

EDIT: the idea is that the fundamental form gives you a "local" version of length, so that, when integrated gives you the length of a curve, in the same sense that you define , when you do Riemann integration, a length element which you integrate to find the overall length of a curve.

EDIT 2: As you said, the metric is a quadratic form, defined on pairs of vector fields. In the differential
geometry version, a quadratic form is usually called a 2-form, which is a bilinear form ( meaning linear
separately in each variable ) defined on a pair of vector fields.

EDIT3: The post may have sprawled out of control, feel free to ask for clarifications.
 
Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top