In trying to understand the exterior or wedge product following Prof. Shifrin's series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, I would like to make sure that the following is correct:

The dual vector space $V^*=(\mathbb R^n)^*$ is a vector space of linear maps $\mathbb R^n \to \mathbb R$ that can be represented by a $1\times n$ matrix (i.e. row vector). It is isomorphic to $\mathbb R^n,$ and forms a vector space, which basis can be expressed as $\mathrm e^1,\mathrm e^2,\dots,\mathrm e^n,$ or as $\mathrm dx_1, \mathrm dx_2, \dots ,\mathrm dx_n,$ such that $\mathrm dx_i(\vec v)=\vec {\mathrm e}_i\cdot \vec v.$

This dual vector space constitutes the specific case of the vector space of alternating multilinear maps, denoted $\Lambda^k(\mathbb R^n)^*$ when $k=1.$ For example, in 3-dimensional Euclidean space, i.e. $n=3,$ a possible form would be $7\mathrm dx_2=7\mathrm dy=7\mathrm e^2,$ and if this is fed a vector $\vec v=\begin{bmatrix}-7&8&2\end{bmatrix}^\top,$ the result will be simply the dot product $\small\begin{bmatrix}0&7&0\end{bmatrix}\begin{bmatrix}-7\\8\\2\end{bmatrix}=56.$

As a vector space, these forms can be added (and scalar multiplied), so that $4\mathrm dx +7 \mathrm dy=\begin{bmatrix}4&7&0\end{bmatrix},$ an expression naturally belonging to the same vector space $\Lambda^1(\mathbb R^3)^*,$ and also accepting vectors in $\mathbb R^3$ to produce a real number through the dot product operation.

The wedge product of two such forms in $\Lambda^1(\mathbb R^3)^*$ will result in an element in $\Lambda^2(\mathbb R^3)^*.$ For example, taking the last element of $\Lambda^1(\mathbb R^3)^*$ in the preceding example, i.e. $\psi =4\mathrm dx_1 +7 \mathrm dx_2,$ and wedging it with $\varphi =2\mathrm dx_1 +5 \mathrm dx_3,$ also in the same vector space $\Lambda^1(\mathbb R^3)^*,$ we obtain

$$\psi\wedge\varphi=4\mathrm dx_1 +7 \mathrm dx_2 \wedge 2\mathrm dx_1 +5 \mathrm dx_3=-14\;\mathrm dx_{12}+20\;\mathrm dx_{13}+35\; \mathbb dx_{23}\tag 1$$

Critically, this is the same as picturing the two forms in $\Lambda^1(\mathbb R^3)^*,$ i.e. $4\mathrm dx_1 +7 \mathrm dx_2=\color{red}{\begin{bmatrix}4&7&0\end{bmatrix}}^\top$ and $2\mathrm dx_1 +5 \mathrm dx_3=\color{blue}{\begin{bmatrix}2&0&5\end{bmatrix}}^\top$ as

$$\begin{bmatrix} \color{red}4&\color{red}7&\color{red}0\\ \color{blue}2& \color{blue} 0 & \color{blue}5 \end{bmatrix}$$

and realizing that $\mathrm dx_{12}$ is simply the determinant of the first two columns (minor of the submatrix):

$$\det{\begin{bmatrix} \color{red}4&\color{red}7\\ \color{blue}2& \color{blue} 0 \end{bmatrix}}=-14$$

and that $\mathrm dx_{13}$

$$\det{\begin{bmatrix} \color{red}4&\color{red}0\\ \color{blue}2& \color{blue} 5 \end{bmatrix}}=20$$

with $\mathrm dx_{23}$

$$\det{\begin{bmatrix} \color{red}7&\color{red}0\\ \color{blue}0& \color{blue} 5 \end{bmatrix}}=35$$

We could express this as

$$\psi\wedge\varphi=\sum \text {minors}_{ij}\; \mathrm d_{ij}$$

The wedge product can be carried out between vector spaces $\Lambda^l(\mathbb R^n)^*\wedge \Lambda^k(\mathbb R^n)^*\in \Lambda^{l+k}(\mathbb R^n)^*.$

This all comes together when two vectors in $\mathbb R^3$ are fed into the expression in Eq (1), for example $\small\vec p =\begin{bmatrix}-1&-\pi&-\sqrt 2\end{bmatrix}^\top$ and $\small\vec q =\begin{bmatrix}10^1&10^2&10^3\end{bmatrix}^\top:$

$$\begin{align}\left(\psi \wedge \varphi\right)[p,q] &= \require{cancel}\cancel {\left(-14\;\mathrm dx_{12}+20\;\mathrm d_{13}+35\; \mathbb d_{23} \right)\,[p,q]}\\&\cancel{=(-14)(-1)(10^2)+(20)(-1)(10^3)+(35)(-\pi)(10^3)} \end{align}$$

Attempt at mending after Professor Shifrin's correction:

$$\begin{align}\left(\psi \wedge \varphi\right)(p,q) &= -14\,\mathrm dx_{12}(p,q) + 20 \,\mathrm dx_{13}(p,q) + 35\, \mathrm dx_{23}(p,q)\\[2ex] &=-14 \,\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}+ 20\, \begin{vmatrix}-1 & 10 \\-\sqrt 2 & 10^3\end{vmatrix} + 35\, \begin{vmatrix}-\pi & 10^2 \\-\sqrt 2 & 10^3\end{vmatrix} \end{align}$$

As for the equivalence to tensor algebra pointed out in the answer

$$\begin{align} \mathrm dx_{12}(p,q)&=\mathrm dx_1 \wedge \mathrm dx_2 (p,q)\\[2ex] &= \left(\mathrm dx_1 \otimes \mathrm dx_2 - \mathrm dx_2\otimes \mathrm dx_1\right)(p,q)\\[2ex] &= (\mathrm dx_1 \otimes \mathrm dx_2 )(p,q)-(\mathrm dx_2\otimes \mathrm dx_1)(p,q)\\[2ex] &=\mathrm dx_1(p)\,\mathrm dx_2(q) - \mathrm dx_2(p) \,\mathrm dx_1(q)\\[2ex] &= (-1)(10^2) - (-\pi)(10)\\[2ex] &=\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}. \end{align}$$

  • 1
    $\begingroup$ Not worth an entire answer, but when you write "dual vector space [ . . .] is a linear map", don't you mean "a vector space of linear maps"? $\endgroup$
    – PersonX
    Apr 3, 2018 at 20:06
  • $\begingroup$ @PersonX Yes, thank you for your help reading this long question. I figured that if nobody points out any major flaws it might even help someone searching the web for this topic. $\endgroup$ Apr 3, 2018 at 20:08
  • $\begingroup$ You have a bunch of typos. You mean to have $dx_{ij}$ everywhere, which of course is (my) shorthand for $dx_i\wedge dx_j$. I also don't particularly like using square brackets around the $p$ and $q$ (as that suggests all sorts of other things in mathematics. An element of $\Lambda^k(\Bbb R^{n*})$ should be applied to an ordered $k$-tuple of vectors. $\endgroup$ Apr 3, 2018 at 21:21
  • $\begingroup$ Thank you for looking at it. The typos you are referring to (please feel free to edit) are in the English explanations (common language) or do you mean that $dx_{ij}$ should have been used instead of $\mathrm dx_{ij}$? $\endgroup$ Apr 3, 2018 at 21:30
  • 1
    $\begingroup$ Oh, I don't care about $d$ versus d. No, you dropped the $x$'s and typed $d_{13}$ and $\mathbb d_{23}$. $\endgroup$ Apr 3, 2018 at 21:44

2 Answers 2


Everything looks fine up until the end, where you should have, for example, $dx_{12}(p,q) = \left|\begin{matrix}-1&10 \\-\pi & 100\end{matrix}\right| = -100+10\pi$. Fitting things into the tensor algebra picture is quite simple: $$dx_{12} = dx_1\wedge dx_2 = dx_1\otimes dx_2 - dx_2\otimes dx_1.$$

  • $\begingroup$ Now, the vectors p and q are being fed into the $\Lambda^2$ form $\psi \wedge \varphi.$ The determinants of the submatrices were already calculated. At that point, I wanted to figure out the expected effect of this multilinear form on the two vectors p and q. Can you please look at it again, and perhaps include the whole calculation so that there is no miscommunication? $\endgroup$ Apr 3, 2018 at 21:34
  • 1
    $\begingroup$ I'm talking about the last line of your computation. Minors show up in two places, both in wedging $\psi$ and $\phi$ together and in evaluating on the vectors. As I typed, by definition, $dx_{12}(p,q)$ computes the $12$-minor of the $3\times 2$ matrix with columns $p$ and $q$. What I'm talking about occurs when you compute $-14 dx_{12}(p,q)$. It should be $-14(-100+10\pi)$. $\endgroup$ Apr 3, 2018 at 21:46

Just some notes on the mechanics of forms - a good place to park some notes I used recently...

First off, regarding the specific example in the OP, it can equivalently be unfolded as follows:

$$\small \psi \wedge \varphi(p,q) = \begin{vmatrix} \psi(p) && \psi(q) \\ \varphi(p) && \varphi(q) \end{vmatrix}= \psi(p)\varphi(q) - \psi(q)\varphi(p)=(-4-7\pi)(20 + 5\cdot 10^3)-(40 + 7\cdot 10^2)(-2-5\sqrt 2)$$

Some more general notes:

An $m$ differential form in $\mathbb R^n$ can be expressed in multi-index notation as

$$\begin{align} \omega &= \sum_I f_I \; dx_I \\ &= \sum_{1 \leq i_1 < i_2 \ \dots <i_m} a_{i_1, \dots, i_m} \; dx_1 \wedge dx_2 \wedge \dots \wedge dx_m \\\\ &I=(i_1, i_2, \dots,i_m) \; \mid\; 1 \leq i_1 < i_2 \ \dots <i_m \leq n \end{align}$$

In $\mathbb R^5$, we can construct $1$-forms using the $dx$ elements of the dual basis $(1,3),(1,2,3,4,5)$ or $(2,4,5):$

\begin{align} \color{blue}{dx_1} && dx_2 &&\color{blue}{dx_3} && dx_4 && dx_5 \\ \color{orange}{dx_1} && \color{orange}{dx_2} && \color{orange}{dx_3} && \color{orange}{dx_4} && \color{orange}{dx_5} \\ dx_1 && \color{red}{dx_2} && {dx_3} && \color{red}{dx_4} && \color{red}{dx_5} && \\ \end{align}

These are not the multi-indices above, which correspond to wedged dual basis elements.

An example of a $1$-forms with these dual basis elements can look like

$$\begin{align}\omega_1 &= -7\,\color{blue}{dx_1} + 3 \,\color{blue}{dx_3}\\ \omega_2 &=-1 \,\color{orange}{dx_1} + \color{orange}{dx_2} + \color{orange}{dx_3}-4\,\color{orange}{d_4} + 5\,\color{orange}{dx_5}\\ \omega_3 &=-2 \,\color{red}{dx_2} + \color{red}{dx_4} - \color{red}{dx_5} \end{align} $$

A $3$-form can be the result of wedging $\omega_1 \wedge \omega_2\wedge \omega_3.$

It acts on three vectors in $\mathbb R^5,$ such as

$$\begin{align}\small v_1 &=\begin{bmatrix}1&&2 &&-1&& 0 &&5\end{bmatrix}\\ v_2 &=\begin{bmatrix}0&&0 &&1&& 10 &&1\end{bmatrix}\\ v_3 &=\begin{bmatrix}7&&1 &&0&& 3 &&2\end{bmatrix} \end{align}$$


$$\omega_1 \wedge \omega_2\wedge \omega_3\;(v_1,v_2,v_3)=\begin{vmatrix}\color{blue}{-10}&&\color{orange}{25} && \color{red}{-9}\\\color{blue}{3}&&\color{orange}{-34} && \color{red}{9}\\\color{blue}{-49}&&\color{orange}{-8} && \color{red}{-1}\end{vmatrix}=3200$$

Trying to express it as in the multi-index notation above invoves distributing the wedged $1$-forms:

$$\begin{align} \omega_1 \wedge \omega_2\wedge \omega_3 &=\Tiny\left( -7\, dx_1+ 3\,dx_3\right)\wedge \left(-1\,dx_1 + dx_2 + dx_3-4\,dx_4 + 5\,dx_5 \right) \wedge \left(-2\,dx_2 + dx_4 - dx_5 \right)\\ &\Tiny=( -7 \, dx_1\wedge dx_2 - 7\, dx_1\wedge dx_3 +28\, dx_1 \wedge dx_4 -35\,dx_1\wedge dx_5 +3 dx_1\wedge dx_3\\ &\Tiny -3\,dx_2\wedge dx_3 - 12\,dx_3\wedge dx_4) + 15 \, dx_3 \wedge dx_5\\ &\Tiny\wedge \Tiny \left(-2\,dx_2 + dx_4 - dx_5 \right)=\\ & \small -8\,dx_1\wedge dx_2 \wedge dx_3\\ & \small + 49\,dx_1\wedge dx_2 \wedge dx_4 \\ & \small - 63\, dx_1\wedge dx_2 \wedge dx_5 \\ &\small -4\, dx_1\wedge dx_3 \wedge dx_4 \\ & \small + 4 \, dx_1\wedge dx_3 \wedge dx_5\\ & \small +7 \, dx_1\wedge dx_4 \wedge dx_5\\ &\small + 21\, dx_2\wedge dx_3 \wedge dx_4\\ &\small - 27\, dx_2\wedge dx_3 \wedge dx_5\\ &\small - 3\, dx_3\wedge dx_4 \wedge dx_5 \end{align}$$

which applied to the vectors above:

$$\Tiny -8 \begin{vmatrix} 1& 2 & -1 \\ 0&0&1\\7&1&0 \end{vmatrix} +49\begin{vmatrix} 1&2&0 \\ 0&0&10\\7&1&3 \end{vmatrix} -63\begin{vmatrix} 1&2&5 \\ 0&0&1\\7&1&2 \end{vmatrix} -4 \begin{vmatrix} 1&-1&0 \\ 0&1&10\\7&0&3 \end{vmatrix} + 4 \begin{vmatrix} 1&-1&5 \\ 0&1&1\\7&0&2 \end{vmatrix} +7 \begin{vmatrix} 1&0&5 \\ 0&10&1\\7&3&2 \end{vmatrix} +21 \begin{vmatrix} 2&-1&0 \\ 0&1&10\\1&0&3 \end{vmatrix} -27 \begin{vmatrix} 2&-1&5 \\ 0&1&1\\1&0&2 \end{vmatrix} -3 \begin{vmatrix} -1&0&5 \\ 1&10&1\\0&3&2 \end{vmatrix}=3200$$

Here is another example:

$$\begin{align}\omega_1 &= dx + dy + dz\\ \omega_2 &=2\, dx - 3 \, dy\\ \omega_3 &=dx + 2\, dz \end{align} $$

A $3$-form can be the result of wedging $\omega_1 \wedge \omega_2\wedge \omega_3.$

It acts on three vectors in $\mathbb R^3,$ such as

$$\begin{align}\small v_1 &=\begin{bmatrix}2&1 &0\end{bmatrix}\\ v_2 &=\begin{bmatrix}-1&3&-2\end{bmatrix}\\ v_3 &=\begin{bmatrix}1&0&1\end{bmatrix} \end{align}$$

$$\omega_1 \wedge \omega_2\wedge \omega_3\;(v_1,v_2,v_3)=\begin{vmatrix}3&&1 && 2\\0&&-11&& -5\\2&&2 && 3\end{vmatrix}=-35$$

If instead we distribute as

$$\small\begin{align} (dx + dy + dz) \wedge (2dx - 3dy) \wedge (dx+2dz) &= (-3\,dx\wedge dy - 2\,dx\wedge dy -2\,dx\wedge dz+3\,dy\wedge dz )\wedge (dx+2dz)\\ &= (-5\,dx\wedge dy -2\,dx\wedge dz+3\,dy\wedge dz)\wedge (dx+2dz)\\ &= -10\, dx\wedge dy \wedge dz + 3 dx \wedge dy \wedge dz \\ &= - 7\, dx\wedge dy \wedge dz \end{align}$$

the same result could now be obtained multiplying...

$$-7 \; \begin{vmatrix}2 & 1 & 0 \\ -1 & 3 & -2 \\ 1 & 0 & 1 \end{vmatrix}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .