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Tensors, the geometric tool that solved Einstein's relativity problem

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Re: Tensors, the geometric tool that solved Einstein's relativity problem

#31

Post by ogogmad »

btilly wrote: post #29
g15jv2dp wrote: post #21

Earlier quoted context omitted.

That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.

You are right about infinite dimensions, wrong about finite dimensions. V and V* are naturally isomorphic for finite dimensions. In finite dimensions, V and V* are isomorphic, but not naturally so. The isomorphism requires additional information. You can specify a basis to get the isomorphism, but many bases will give the same isomorphism. The exact amount of information that you need is a metric. If you have a metri…

You need to correct the 2nd sentence to say that V and V** are naturally isomorphic. V and V* are only unnaturally isomorphic. All of this holds only in finite dimensions, of course.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#32

Post by [deleted] »

senderista wrote: post #26
g15jv2dp wrote: post #21

Earlier quoted context omitted.

That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.

No, the isomorphism between V and V** (for finite-dimensional V) is canonical. The canonical isomorphism T:V->V** is easy to construct: map a vector v in V to the element of V** which takes an element w from V* and applies it to v: T(v)(w) = w(v).

[deleted]

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#33

Post by andrewla »

dannymi wrote: post #16
max_likelihood wrote: post #3

I've always thought the use of "Tensor" in the "TensorFlow" library is a misnomer. I'm not too familiar with ML/theory, is there a deeper geometric meaning to the multi-dimensional array of numbers we are multiplying or is "MatrixFlow" a more appropriate name?

In the first example on https://www.tensorflow.org/api_docs/python/tf/math/multiply you can see that they use the Hadamard product (not the matrix product): x = tf.constant(([1, 2, 3, 4])) tf.math.multiply(x, x) I could stop right here since it's a counterexample to x being a matrix (with a matrix product defined on it; P.S. try tf.matmul(x, x)--it will fail; there's no .transpose either). But that's only technically…

In TensorFlow the tf.matmul function or the @ operator perform matrix multiplication. Element-wise multiplication ends up being useful for a lot of paralellizable computation but should not be confused with matrix multiplication.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#34

Post by g15jv2dp »

senderista wrote: post #26
g15jv2dp wrote: post #21

Earlier quoted context omitted.

That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.

No, the isomorphism between V and V** (for finite-dimensional V) is canonical. The canonical isomorphism T:V->V** is easy to construct: map a vector v in V to the element of V** which takes an element w from V* and applies it to v: T(v)(w) = w(v).

GP is giving you an element of V**. You want to turn it into a vector. To do that, please make the inverse isomorphism explicit without using a basis. I'll wait...

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#35

Post by g15jv2dp »

btilly wrote: post #29
g15jv2dp wrote: post #21

Earlier quoted context omitted.

That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.

You are right about infinite dimensions, wrong about finite dimensions. V and V* are naturally isomorphic for finite dimensions. In finite dimensions, V and V* are isomorphic, but not naturally so. The isomorphism requires additional information. You can specify a basis to get the isomorphism, but many bases will give the same isomorphism. The exact amount of information that you need is a metric. If you have a metri…

You have a typo in your first line, and I answered a sibling comment about that. Metrics are irrelevant to the discussion (and I presume you wanted to write "norm" instead of "metric").

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#36

Post by thechao »

SkyBelow wrote: post #20
andrewla wrote: post #10

Earlier quoted context omitted.

Slight disagree here -- matrices are enough for transformations in 2, 3, 4, and 100 dimensions. Tensors are not arrays with more rows and columns; they are higher dimensional objects -- more indices, not greater range of indices.

Is a tensor higher dimension, or is it the generalized form of the structure encompassing all of it (individual numbers (0 dimensions), vectors (1 dimension), matrixes (2 dimensions), and so on)? Kind of like how an n-sphere describes circles and spheres for n equal to 2 or 3.

> (individual numbers (0 dimensions), vectors (1 dimension), matrixes (2 dimensions), and so on)

You're using the word "dimension" in two (distinct) ways. Instead, use the word "rank":

> (individual numbers (0 rank), vectors (1 rank), matrixes (2 rank), and so on)

Now, we can talk about a 4-dimensional rank-1 tensor, e.g., a 4-element vector.

Now, think about a 4x4 matrix: if we multiply the matrix by a 4-vector, we get a 4-vector out: in some ways, the multiplication has "eaten" one of the ranks of the matrix; but, the dimension of the resulting object is the same. If we had a 3x2 matrix, and we multiplied it by a 3-vector, then both the rank has changed (from 2 to 1) and the dimension has changed (from 3 to 2).

A tensor has any number of rank.

More importantly, the ranks of a tensor come in two "flavors": a vector and a one-form. The concepts are pretty darn general, but one way to get a feel for how they're related is that the transpose of a vector can be its dual. This gets into things like pre- and post- multiplication; or, whether we 'covary' or 'contravary' with respect to the tensor.

Frankly, tensor products are a beast to deal with, mechanically, so the literature mostly deals with them as opaque objects. The modern tensor software libraries and high performance computing has seen a sea-change in the use of GR.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#37

Post by btilly »

g15jv2dp wrote: post #35
btilly wrote: post #29

Earlier quoted context omitted.

You are right about infinite dimensions, wrong about finite dimensions. V and V* are naturally isomorphic for finite dimensions. In finite dimensions, V and V* are isomorphic, but not naturally so. The isomorphism requires additional information. You can specify a basis to get the isomorphism, but many bases will give the same isomorphism. The exact amount of information that you need is a metric. If you have a metri…

You have a typo in your first line, and I answered a sibling comment about that. Metrics are irrelevant to the discussion (and I presume you wanted to write "norm" instead of "metric").

Yes, I had a typo.

As for why I said metric, see https://en.wikipedia.org/wiki/Metric_tensor. Which is technically a concept from differential geometry rather than linear algebra. But then again, tensors are literally the topic that started this. And it is only in differential geometry that I've ever cared about mapping from V to V*.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#38

Post by mr_mitm »

senderista wrote: post #2

If you have any linear algebra background, then the definition of a tensor is straightforward: given a vector space V over a field K (in physics, K = R or C ), a tensor T is a multilinear (i.e. linear in each argument) function from vectors and dual vectors in V to numbers in K . That's it! A type (p, q) tensor T takes p vectors and q dual vectors as arguments ( p+q is often called the rank of T but is ambiguous comp…

Yes, very simple, except that when physicists say "tensor", they mean tensor fields, on smooth, curved manifolds, in at least four dimensions, often with a Lorentz metric. Things stop being simple quickly.

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#39

Post by g15jv2dp »

btilly wrote: post #37
g15jv2dp wrote: post #35

Earlier quoted context omitted.

You have a typo in your first line, and I answered a sibling comment about that. Metrics are irrelevant to the discussion (and I presume you wanted to write "norm" instead of "metric").

Yes, I had a typo. As for why I said metric, see https://en.wikipedia.org/wiki/Metric_tensor . Which is technically a concept from differential geometry rather than linear algebra. But then again, tensors are literally the topic that started this. And it is only in differential geometry that I've ever cared about mapping from V to V*.

There are no manifolds at all in this discussion... What are you even talking about? Just stringing words you vaguely know together??

Re: Tensors, the geometric tool that solved Einstein's relativity problem

#40

Post by klodolph »

will-burner wrote: post #6
senderista wrote: post #2

If you have any linear algebra background, then the definition of a tensor is straightforward: given a vector space V over a field K (in physics, K = R or C ), a tensor T is a multilinear (i.e. linear in each argument) function from vectors and dual vectors in V to numbers in K . That's it! A type (p, q) tensor T takes p vectors and q dual vectors as arguments ( p+q is often called the rank of T but is ambiguous comp…

The definition may be simple, but it's not very concrete and I'd argue that makes it not strait forward. While examples of vector spaces can be very concrete (think R, R^2, R^30), I struggle to think of a concrete example of a multilinear function from vectors and dual vectors in V to numbers in K. On top of that when working with tensors, you don't usually use the definition os a multilinear function at least as far…

I think part of this is “if you have a linear algebra background”. There are a few different explanations of tensors, and different explanations make sense for different people.
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