# Tensor exercise multilinear algebra

Determine which of the following are tensors on $\mathbb{R}^4$, and express those in terms of elementary tensors $$f(x,y,z) = 3x_1y_2z_3 - x_3 y_1 z_4.$$

The solution say

My questions

1. What does tensors on $\mathbb{R}^4$ mean? It means it takes on a vector with four components in real-valued right? But my function has $9$ variables, namely $f(x_1,x_2,x_3,y_1,y_2,z_1,z_2,z_3,z_4)$. Unless it actually has $12$ variables, $f(x_1,x_2,x_3,x_4,y_1,y_2, y_3, y_4,z_1,z_2,z_3,z_4).$

2. Here is my work trying to verify the elementary tensor product is correct. I know that $$\omega^i (\mathbf{v}) = v_i$$

\begin{align} f(x_1,x_2,x_3,x_4,y_1,y_2, y_3, y_4,z_1,z_2,z_3,z_4). &= (3\omega^1 \otimes \omega^2 \otimes \omega^3 - \omega^3 \otimes\omega^1 \otimes\omega^4 )(x_1,x_2,x_3,x_4,y_1,y_2, y_3, y_4,z_1,z_2,z_3,z_4).\\ &=3\omega^1 (\mathbf{x,y,z}) \omega^2(\mathbf{x,y,z}) \omega^3((\mathbf{x,y,z}) - \omega^3 (\mathbf{x,y,z}) \omega^1(\mathbf{x,y,z}) \omega^4(\mathbf{x,y,z}) \end{align}

If I work backwards I can figure out what the pattern looks like. But I don't understand why once we evaluate $\omega^1$ on $\mathbf{x}$, the $\omega^2$ that comes after evaluates on $\mathbf{y}$?

I am very new to this. My book doesn't define nor explain what it means to evaluate on two vectors. I don't even know what is the order of the tensors $\omega^i$s.

• Curiosity: what book are you using? I will begin to study this subject soon. Thank you (: – Ivo Terek May 31 '14 at 2:38
• Right now, I am looking at Munkres. – Hawk May 31 '14 at 3:15

O.K, here is a better answer; I think the other answer is also informative ( actually, I put in a lot of time into writing it , so I don't want to delete it :) ):

1) A tensor f in $\mathbb R^4$ is a function defined on $\mathbb R^4$ that is linear on each variable separately, i.e., f is linear in each variable when other variables are held constant. In our case, the variables in $\mathbb R^4$ are $x=(x_1,x_2,..,x_4),...,z= (z_1, z_2, z_3, z_4)$. I think what is confusing is the way the function is expressed, as if it was a function defined on $\mathbb R^9$ , or so. I think this can be remedied by using the projection maps $\pi_j ; j=1,2,3,4$, with $\pi_j(x):=x_j$. Then the map $$f(x,y,z)=3x_1y_2z_3 -x_3y_1z_4$$ , can be written in terms of $(x,y,z)$ alone (as f(x,y,z) ought to, since by definition it only depends on these 3 variables ) by $$f(x,y,z) = 3\pi_1(x)\pi_2(y)\pi_3(z)-\pi_3(x)\pi_1(y)\pi_4(z)$$.

This is a (composite) function of the three variables $(x,y,z)$ , just like, say , $cosx^2+tany$ is a function of two variables.

We know check that this f is a tensor, i.e., that f is linear on each of the variables $(x,y,z)$ ( notice $f$ is defined on a subspace of $\mathbb R^4$ consisting of the first 3 axes.). So we need to check, for all constants $c,d,k$ :

i ) $f(cx,y,z)=cf(x,y,z); f(x,dy,z)=df(x,y,z); f(x,y, kz)=kf(x,y,z)$.

Let me just do the part for $x$; the rest will likely be clear (let's forget the issue with the projections for now ):

$$f(cx,y,z):=f((cx_1,cx_2, cx_3, cx_4),(y_1,..,y_4),(z_1,..,z_4)=3(c_1)y_2 z_3-(cx_3)y_1z_4=c(x_1y_2z_3-x_3y_1z_4)=cf(x,y,z)$$

Now we need to check, for $x'=(x_1',x_2', x_3', x_4')$:

ii)$$f(x+x',y,z)=f(x,y,z)+f(x',y,z)$$ (same for the $y,z$ variables). We have:

$$f(x+x',y,z)=3(x_1+x_1')y_2z_3-(x_3+x_3')y_1z_4=3x_1y_2z_3+ 3x_1'y_2z_3 -(x_3y_1z_4)-(x_3')y_1z_4$$ , which you can verify equals $f(x,y,z)+f(x',y,z)$ . A similar argument shows $f$ is linear with respect to $y,z$.

Now, to verify the correctness of the representation: $$3w^1 \otimes w^2 \otimes w^3 -w^3 \otimes w^1 \otimes w^4 :$$ , using the definition $w^i(e_j)= \delta_i^j$ , where $e_i$ ; i=1,2,3,4 is the "standard" basis $e_1=(1,0,0,0),..,e_4=(0,0,0,1)$.

In this basis, a point $x=(x_1,x_2,x_3, x_4)$ has a representation $$x_1e_1+x_2e_2+x_3e_3+x_4e_4$$ . Also, the tensor operation $w^i \otimes w^j \otimes w^k$ is defined by $$w^i \otimes w^j \otimes w^k (a,b,c):=w^i(a)w^j(b)w^k(c)$$ , where $a,b,c$ are vectors.

We then have :

$$3w^1 \otimes w^2 \otimes w^3 ((x_1e_1+...x_4e_4),...,(z_1e_1,...,z_4e_4)=3w^1(x_1e_1,..x_4e_4)\otimes w^2(y_1e_1,..,y_4e_4) \otimes w^3(z_1e_1,..,z_4e_4)=3x_1y_2z_3$$ . Try verifying the other component $w^3 \otimes w^1 \otimes w^4$.

• I think you mean a subset of $\mathbb{R}^4$, not a subspace. Also how do we know (my book has a proof about this, but they didn't explain how they came up with this definition) $$w^i \otimes w^j \otimes w^k (a,b,c):=w^i(a)w^j(b)w^k©$$ – Hawk May 31 '14 at 8:14
• If you refer to $(x,y,z,0) \subset \mathbb R^4$ , this is a subspace. In my understanding, the formula you ask about is a definition. – user99680 May 31 '14 at 8:21
• @sidht: Maybe this is the answer you wanted: we are working with the tensor algebra of $V$. We know by construction that the tensor product of a k-tensor and an n-tensor is a (k+n)-tensor. We then extend by linearity: en.wikipedia.org/wiki/Tensor_algebra – user99680 May 31 '14 at 19:06
• @sidht: Maybe more elementarily, k-tensors act on k-vectors, by construction. In this case, the $w^k's$ are 1-tensors, so the act on vectors. Each of $w^i, w^j, w^k$ are 1-tensors, and so each acts on the vectors, or "1-vectors" a,b,c. ^ – user99680 Jun 2 '14 at 0:37