# GR Math: how does tensor linearity work?

1. Mar 23, 2009

### nonne

So I'm reading these notes about differential geometry as it relates to general relativity. It defines a tensor as being, among other things, a linear scalar function, and soon after it gives the following equation as an example of this property of linearity:

T(aP + bQ, cR + dS) = acT(P, R) + adT(P, S) + bcT(Q, R)+bdT(Q, S)

where T is the tensor function, P, Q, R, and S are vectors, and a, b, c, and d are scalar coefficients.

Now I can follow the above leap from left hand side to right hand side as far as:

T(aP + bQ, cR + dS) = T(aP, cR + dS) + T(bQ, cR + dS) = T(aP, cR) +T(aP, dS) + T(bQ, cR) + T(bQ, dS)

but I don't quite understand the reasoning behind how the coefficients get outside of the function brackets. Somehow I managed to get a bachelors in physics without ever taking a single linear algebra course, so I'm a little bit stumped.

Can anyone here give me a hand with this? Any help would be greatly appreciated.

2. Mar 23, 2009

### slider142

A linear operator T satisfies T(sa + b) = sT(a) + T(b) where s is a number. Tensors are multilinear operators (linear in each argument).

3. Mar 23, 2009

### nonne

Yeah, no, I understand that bit, but does that also mean that:

T(sa, tb) = stT(a, b)

where s and t are scalars, for a linear operator T with multiple arguments?

4. Mar 24, 2009

### slider142

It is linear in each argument: T(sa, tb) = sT(a, tb) = stT(a, b).

5. Apr 1, 2009

### VKint

It may or may not be helpful to think of a (0,2) tensor as a regular linear map from a vector space $$V$$ to the space of all linear maps from $$V$$ to $$\mathbb{R}$$. Such a statement sounds convoluted, but you've actually encountered this kind of thing before: Given a function $$f : \mathbb{R}^n \to \mathbb{R}^m$$, the full derivative of $$f$$ is a mapping $$D : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R}^m)$$, where $$L(\mathbb{R}^n,\mathbb{R}^m)$$ denotes the space of all linear maps $$\mathbb{R}^n \to \mathbb{R}^m$$ (i.e., the space of all $$m \times n$$ matrices), such that $$D(p)$$ is the Jacobian of f at $$p$$. However, the map $$D$$ isn't necessarily linear (although the Jacobian at any given point certainly is).

In the same way, a (0,2) tensor can be thought of as a function $$T: V \to L(V,k)$$, where $$k$$ is the base field. (The space $$L(V,k)$$ is better known as the dual vector space, $$V^{*}$$.) Specifically, if $$T$$ is a (0,2) tensor, there are two ways to define the map $$T(\mathbf{v}) : V \to V^{*}$$: Either $$T(\mathbf{v}) : \mathbf{w} \mapsto T(\mathbf{v},\mathbf{w})$$, or $$T(\mathbf{v}) : \mathbf{w} \mapsto T(\mathbf{w},\mathbf{v})$$. The defining property of such a tensor is that, considered as a map $$V \to V^{*}$$, it is linear (regardless of which ordering you choose). Likewise, higher-rank tensors can be thought of as linear maps from some space $$V^{*} \otimes \ldots \otimes V^{*} \otimes V \otimes \ldots \otimes V$$ to $$V^{*}$$ or $$V$$. Using this property of tensors, it is relatively trivial to show that $$T(a\mathbf{v},b\mathbf{w}) = abT(\mathbf{v}, \mathbf{w})$$.

6. Apr 1, 2009

### VKint

It may or may not be helpful to think of a (0,2) tensor as a regular linear map from a vector space $$V$$ to the space of all linear maps from $$V$$ to $$\mathbb{R}$$. Such a statement sounds convoluted, but you've actually encountered this kind of thing before: Given a function $$f : \mathbb{R}^n \to \mathbb{R}^m$$, the full derivative of $$f$$ is a mapping $$D : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R}^m)$$, where $$L(\mathbb{R}^n,\mathbb{R}^m)$$ denotes the space of all linear maps $$\mathbb{R}^n \to \mathbb{R}^m$$ (i.e., the space of all $$m \times n$$ matrices), such that $$D(p)$$ is the Jacobian of f at $$p$$. However, the map $$D$$ isn't necessarily linear (although the Jacobian at any given point certainly is).

7. Apr 1, 2009

### CFDFEAGURU

Tensors are a machine that you insert vectors into and obtain a real number. It is that simple.

8. Apr 1, 2009

### VKint

WTF?...Sorry for the double post.