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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), m

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.

andrewla 2y ago
The problem is that I'm using dimensional in two different senses. A vector with N elements, for example, is a one-dimensional array of numbers operating on N dimensional spaces, and so on for matrices and tensors. If you're doing graphics programming you're operating in three and four dimensional spaces mostly (four dimensional being projective spaces), because that's the space you're trying to render in two dimensions. But you'll rarely need anything higher than a matrix (a two-dimensional data structure) for operations on that space. If you're doing physics you're operating in three, four, and infinite dimensional spaces, mostly. And you'll routinely use higher data structures -- even things like moment of inertia for rigid bodies can't really be described without rank 3 tensors (a three-dimensional data structure). In statistics and machine learning, you're operating in very high dimensional spaces, and will find yourself using non-square tensors especially (in the other areas everything will be square). The data structures will generally be high dimensional as well, but usually just a function of model complexity; so maybe 4 or 5 dimensional data structures.
enugu 2y ago
Any given tensor has a type (p,q). Say d=p+q. Then d is the 'dimension' of the tensor in the sense that you would need nxnx...n=n^d numbers (think of an nxnx...n array) to describe the tensor where n is the dimension of the underlying vector space. > or is it the generalized form of the structure encompassing all of it...Kind of like how an n-sphere If you are asking whether there are examples of tensors parametrized by an integer d, you can cook up examples - like the (d,0) tensor whose input is a sequence of d vectors and just adds up all the components with respect to some basis in each slot and then adds this number across all the d slots. But just like a n-spere is a special example, of a polynomial in n variables, the above tensor is a specific example - it is the tensor where all the components are 1 in the higher dimensional array. Sometimes, one considers the algebra of tensors across all dimensions like the symmetric algebra or exterior algebra simultaneously (where there is multiplication operation between tensors of different dimensions), but that might not be what you were asking about.
thechao 2y ago
> (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.

andrewla 2y ago

The problem is that I'm using dimensional in two different senses. A vector with N elements, for example, is a one-dimensional array of numbers operating on N dimensional spaces, and so on for matrices and tensors. If you're doing graphics programming you're operating in three and four dimensional spaces mostly (four dimensional being projective spaces), because that's the space you're trying to render in two dimensions. But you'll rarely need anything higher than a matrix (a two-dimensional data structure) for operations on that space. If you're doing physics you're operating in three, four, and infinite dimensional spaces, mostly. And you'll routinely use higher data structures -- even things like moment of inertia for rigid bodies can't really be described without rank 3 tensors (a three-dimensional data structure). In statistics and machine learning, you're operating in very high dimensional spaces, and will find yourself using non-square tensors especially (in the other areas everything will be square). The data structures will generally be high dimensional as well, but usually just a function of model complexity; so maybe 4 or 5 dimensional data structures.

enugu 2y ago

Any given tensor has a type (p,q). Say d=p+q. Then d is the 'dimension' of the tensor in the sense that you would need nxnx...n=n^d numbers (think of an nxnx...n array) to describe the tensor where n is the dimension of the underlying vector space. > or is it the generalized form of the structure encompassing all of it...Kind of like how an n-sphere If you are asking whether there are examples of tensors parametrized by an integer d, you can cook up examples - like the (d,0) tensor whose input is a sequence of d vectors and just adds up all the components with respect to some basis in each slot and then adds this number across all the d slots. But just like a n-spere is a special example, of a polynomial in n variables, the above tensor is a specific example - it is the tensor where all the components are 1 in the higher dimensional array. Sometimes, one considers the algebra of tensors across all dimensions like the symmetric algebra or exterior algebra simultaneously (where there is multiplication operation between tensors of different dimensions), but that might not be what you were asking about.

thechao 2y ago

> (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.