don't want my words on PF
Vectors vi (i=1,2,3,...) in Rn are independent iffOriginally posted by lethe
oh and i assume that you know what it means for vectors to be linearly independent,
A set of vectors (v1,v2,v3,...) is a basis for a vector space V iffand what a basis of a vector space is.
I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.Originally posted by lethe
now, hopefully we are all pretty comfortable with what a normal euclidean vector is. it s basically just an arrow between two points. it has a magnitude and a direction. right?
yes, i do want to de-emphasize the notion of a vector as an arrow with direction. in the next post to come, i will write the definition of an abstract vector space, and ask that the reader abandon any preconceptions about vectors as arrows, and think of a vector as mathematical object obeying certain alebraic rules.Originally posted by Tom
I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.
well, i m not sure that i want to emphasize a definition that relies on transformations, we want to delay any introduction of coordinates and metrics/inner products as much as possible. orthogonality relies on the metric, so i don t want to talk about it. transformation rules of vectors rely on the introduction of coordinates on the manifold, so i don t want to talk about that either, at least to start. i want to define, e.g. tangent vectors to a manifold without any reference to local coordinates, and then derive the transformation rule for coordinate transformations, including rotations.
When something is said to be a "vector", one has to specify a set of transformations with respect to which that object is a vector. In the case of Euclidean 3-space, that set of transformations is rotations and parity.
Definition: A vector in Euclidean 3-space (E3) is a mathematical object that transforms under rotations R and parity Π as follows.
where R is an orthogonal matrix (RTR=1). Orthogonality is important because the norm of the vector must be preserved under the rotation.
Explicitly, we must have:
in terms of row and column vectors (vT and v, respectively):
For the equality of the inner products to hold, we can see that we must have RTR=1.
IMO, when vectors are defined in terms of transformations, the extension to other vector spaces and to higher rank tensors in the same vector space is most natural.
Lethe, if you don't mind, could you wait to post the next section for another day? I would like to pick a few exercises out of my linear algebra to reinforce this stuff.
Interesting; I was going to suggest exactly the opposite; I was going to suggest that students think of vectors as bound to a point in space and having a direction and magnitude, to try and quell the notion of a vector as a displacement, and to emphasize that we cannot slide the arrows around like we can in Euclidean space.I don't know if this is going to be an issue with what you are going to bring up later, but when I teach special relativity I try to get the students to stop thinking of vectors in this way, because the "magnitude and direction" definition of a vector is only good for Euclidean space.
[1 2 0|0] [V|[b]0[/b]]=[2 -1 5|0] [3 4 2|0]
[1 2 0|0] [V|[b]0[/b]]=[0 -5 5|0] [0 0 0|0]
[1 -2 1| a] [A|[b]v[/b]]=[-1 3 2| b] [0 1 4| c]
[1 0 0|10a+9b-7c] [A|[b]v[/b]]=[0 1 0|4a+4b-3c] [0 0 1|-a-b+c]
should we let someone solve your exercise before we move on?Originally posted by Tom
Lethe, that's all I wanted to say. The floor is yours.
well, this is supposed to be a suicidal crash course in linear algebra, and cover all the prerequisites. i don t know if that is too ambitious a hope, but at least you re trying. that is promising.Originally posted by gnome
Yikes, lethe, is this thread intended for those of us who haven't taken linear algebra yet, or is that another prerequisite that you neglected to mention? I would like to try to follow it, but it looks like it will require a fair amount of "outside" reading.
OK, yes, a plane is flat 2-dimensional space. an example of a non-flat 2-dimensional space would be the surface of a sphere, or torus (doughnut).
Please try to clarify the concept of a manifold. You said that "a manifold is just a space that is not necessarily flat." I guess that doesn't really help me until I fully understand the concept of a flat space. I mean, it's clear enough that a plane is a 2 dimensional flat space (I hope), but what is a flat 3-space, a flat 4-space, etc.?
not much. for example, the flat space is itself a manifold. so the line, and the plane, those are both manifolds. but a manifold is not necessarily flat. so the circle and the sphere are also manifolds, but they are not flat spaces. what makes them a manifold is that if you are standing very very close to a sphere, and you forgot your glasses, and you don t look around you, you re just looking at one point on the sphere, right on top of your nose, then it will look like it is flat space, and you might not realize that it is actually a sphere.
The examples you gave of curved lines and curved surfaces don't really convey the essence of "manifoldness" (whatever that is). Presumably a space is not necessarily flat. So what distinguishes a manifold from a space?
hmmm.... well, i might introduce orthogonality at some point in this thread. i m not sure. rotations and parity probably not. oh i see, you re refering to the stuff tom was saying to explain what a vector is? i don t think that stuff is too important for now, but there is some disagreement as to what is the best way to introduce vectors.
what course would cover rotations, parity and orthogonality? I haven't come across any of these terms before (at least not in this context). Should I be trying to read about them now, or is that unnecessary?
OK, so let me take another stab at is. do you know what a vector is? the starting place to understand vectors is arrows in R3. they have magnitude and direction. you can rotate them, and reflect them. you can also add them, and scale them.
Also, your summary of the properties that a set of vectors must have to be called a vector space is clear enough, but can you define or explain the concept of a vector space in words?
i am copy-pasting from the other forum, so yes, it will be identical. when i write new entries, i will put them on both threads as well, but obviously discussions/question-answers and such will not be the same.Originally posted by On Radioactive Waves
lethe, will this basicly be the same as that other thread or have you revised it at all?
not a problem. it doesn t have to look flat to everyone, to be a manifold, it only has to look flat to someone who is really really close. and how close you would have to be might depend on how curvy the space is. a really sharply turning curve only looks flat if you re super close, whereas a very broadly turning curve, you don t have to be so close to.Originally posted by gnome
If a manifold is simply a space that either is actually flat or is curvy but up close looks flat, then a Lilliputian and a Brobdingnagian might disagree as to whether a particular space is a manifold. That's not a problem?
that s right. nor do vector dot products. however, we will meet those beasties eventually. a vector dot product is an example of something called an inner product. a vector space that has this additional structure is called an inner product space. once you have an inner product, you can use words like orthogonal, to describe two perpendicular vectors (two vectors are orthogonal iff their inner product is zero), but not before then.
As to vectors, you said "nowhere does a vector space allow you to multiply 2 vectors". So, the vector cross-product doesn't apply to this discussion?
You're talking about "extrinsic" curvature which describes how a surface is embedded in a higher dimensional space. It is the type of curvature in terms of which we ordinarily perceive and describe shapes. However, it's really a surface's "intrinsic" curvature that's of interest here.Originally posted by lethe
what do i mean by curvy (non-flat)? what is an example of a curvy 3-dimensional space? well this is a bit tricky to explain. recall that my examples of non-flat 2-dimensional spaces were surfaces drawn in R 3. it is the case that you always need to draw your space in some R n, where n is more than the dimension of your space. the reason is, that the space neds extra room to bend. like when you bend the line into a circle, you need the second dimension to bend into, even though your space (the circle) is only 1-dimensional. when you bend a plane into a sphere, you need the third dimension to bend around, even though the space (the sphere) is only 2-dimensional. so non-flat 3-dimensional space can only fit in some R nif n is 4 or more.
well, intrinsic curvature is certainly more important when you re doing geometry. but so far, in this thread, we have not introduced any metric, so curvature, either intrinsic or extrinsic, is not defined. really, i was just trying to give a layman description of what it means for a higher dimensional space to be flat or not flat. just an intuitive picture.Originally posted by jeff
You're talking about "extrinsic" curvature which describes how a surface is embedded in a higher dimensional space. It is the type of curvature in terms of which we ordinarily perceive and describe shapes. However, it's really a surface's "intrinsic" curvature that's of interest here.
It's simply wrong - whatever the theme of this thread - to describe "what it means for a higher dimensional space to be flat or not flat" in terms of it's embedding in a higher dimensional space. As you pointed out, surfaces can have the same intrinsic curvature but different extrinsic curvatures.Originally posted by lethe
i was just trying to give a layman description of what it means for a higher dimensional space to be flat or not flat. just an intuitive picture.
A metric is not needed to distinguish between surfaces that are flat and curved. All one needs is the idea of parallel transport which requires only a connection to define.Originally posted by lethe
...we have not introduced any metric, so curvature, either intrinsic or extrinsic, is not defined.
yes, perhaps. but you know, i think the notion of intrinsic geometry takes a little while to develop, whereas anyone can picture a "curvy" embedding. and since i m not going to do anything with metrics or connections in this thread (except possibly the hodge star operator), then i didn t think it was the right place for that.Originally posted by jeff
It's simply wrong - whatever the theme of this thread - to describe "what it means for a higher dimensional space to be flat or not flat" in terms of it's embedding in a higher dimensional space. As you pointed out, surfaces can have the same intrinsic curvature but different extrinsic curvatures.
yeah, well you know what? just like there is no metric yet, there is also no connection yet.
True, but one doesn't need the metric to distinguish between surfaces that are flat and curved. All one needs is the idea of parallel transport which requires only a connection to define.