0
$\begingroup$

My question relates to this reply on math.stackexchange.

More precisely, I am wondering about the following sequence of expressions involving elements of an invertible square matrix $M$ and a pair of Levi-Civita symbols.

First, it is known that determinant $\det M$ can be expressed as $$\det(M)=\tfrac 1{n!}\varepsilon^{i_1i_2\dots i_n}\varepsilon_{j_1j_2\dots j_n}M^{j_1}_{\quad i_1}M^{j_2}_{\quad i_2}\dots M^{j_n}_{\quad i_n}.$$

Second, the formula for the adjugate matrix is very similar to the previous one: $$\mathrm{adj}(M)^{a}_{\; {b}}=\tfrac 1{(n-1)!}\;\varepsilon^{ai_2\dots i_n}\varepsilon_{bj_2\dots j_n}M^{j_2}_{\quad i_2}\dots M^{j_n}_{\quad i_n}$$ and it has a particularly neat interpretation in terms of the inverse $M^{-1}$ and of determinant $\det M$, namely $$(M^{-1})^a_{\; b}=\frac{\mathrm{adj}(M)^a_{\; b}}{\det(M)}.$$ All of the above formulas can be found in various books on differential geometry.

It seems natural to me to construct a `next term' in sequence to the two above functions on $M$: one which further extends the formulas for $\det$ and $\mathrm{adj}$ as $$\mathrm{fun}(M)^{a_1 a_2}_{\; \; \; \; b_1 b_2}=\tfrac 1{(n-2)!\times 2!}\;\varepsilon^{a_1 a_2 i_3\dots i_n}\varepsilon_{b_1 b_2 j_3\dots j_n}M^{j_3}_{\quad i_3}\dots M^{j_n}_{\quad i_n},$$ which however does not seem to be considered in the differential geometry textbooks I checked.

I am trying to understand whether $\mathrm{fun}(M)^{a_1 a_2}_{\; \; \; \; b_1 b_2}$ as defined above can be related in some easy way to matrix inverse/power/determinant, in a spirit similar to the case of the adjugate matrix. My hunch is that two copies of $M^{-1}$ should somehow be involved, but I keep failing to connect the dots.

$\endgroup$
5
  • $\begingroup$ Nice question. We may have to clean up the factors in front of the $\varepsilon$s a bit but I think what happens is that $\operatorname{fun}(M)$ is a fourth order tensor that gives the determinant when contracted twice with $M$ and the adjugate when contracted only once. So, modulo factors, and in coordinate free notation: $$ M^{-1}=\frac{\operatorname{fun}(M)(M)}{\operatorname{fun}(M)(M,M)}\,. $$ $\endgroup$
    – Kurt G.
    Commented Feb 9, 2023 at 6:33
  • $\begingroup$ Thank you! Yes, this looks legit. Although this is not quite what I was hoping for, your comment helps me express more sharply what I am wondering about. Namely, the following multiplication of the adjugate of $M$ with a rank-1 vector results in a simple formula: $\textrm{adj}(M)^a_{\;b}v^b\sim (\det M)M^{-1}v$ in terms of an (inverse) power and determinant of $M$. Could there be a similar kind of formula in terms of powers & determinant of $M$ for the action $\textrm{fun}(M)^{ab}_{\;\;cd} X^{cd}$ on a rank-2 tensor? Perhaps sth where $X$ is multiplied by $M^{-1}$ in both indices $c$ and $d$? $\endgroup$ Commented Feb 9, 2023 at 12:56
  • $\begingroup$ Here a conjecture after playing Mathematica (checked for $m=1,2$, $n=3,4$). Let $\textrm{adj}_m(M)^{a_1\ldots a_m}_{\qquad b_1\ldots b_m}=\frac{1}{(n-m)!m!}\epsilon^{a_1\ldots a_m i_{m+1}\ldots i_n}_{\qquad \qquad \quad b_1\ldots b_m j_{m+1}\ldots j_n}M^{j_{m+1}}_{\qquad i_{m+1}}\cdots M^{j_n}_{\quad i_n}$ be the $m$th adjugate of $M$. Then for $m$-form $X$ it holds that $\textrm{adj}_m(M)^{a_1\ldots a_m}_{\qquad b_1\ldots b_m}X^{b_1\ldots b_m}=(\det M)(M^{-1})^{a_1}_{\;\;b_1}\cdots(M^{-1})^{a_m}_{\;\;b_m}X^{b_1\ldots b_m}$. But I fail to prove it generally. Could my conjecture be true? $\endgroup$ Commented Feb 10, 2023 at 22:58
  • $\begingroup$ Why not? It looks like you invented $\textrm{adj}_m(M)^{a_1\ldots a_m}_{\qquad b_1\ldots b_m}$ to exactly fulfil that purpose. $\endgroup$
    – Kurt G.
    Commented Feb 11, 2023 at 6:39
  • $\begingroup$ Well, the conjectured relation was really not obvious to me! I simply followed the indices where they led me (in some application to a physics problem) and then became stuck at trying to simplify some very similar relation. But I believe I have now finally managed to connect the dots, and typed a proof in an answer below (I hope I didn't miss anything). $\endgroup$ Commented Feb 13, 2023 at 16:44

1 Answer 1

0
$\begingroup$

The relevant pattern here is perhaps the following.

If the adjugate matrix is multiplied by a degree-1 tensor $\alpha$, one obtains (from the identity mentioned in the question) that $$\textrm{adj}(M)^a_{\phantom{a}b} \alpha^b = (\det M) (M^{-1})^a_{\phantom{a}b} \alpha^b.$$ Similarly, multiplication of $\textrm{fun}(M)^{a_1 a_2}_{\phantom{a_1 a_2}b_1 b_2}$ by a degree-2 skew-symmetric tensor $\beta$ results in $$\textrm{fun}(M)^{a_1 a_2}_{\phantom{a_1 a_2} b_1 b_2} \beta^{b_1 b_2} = (\det M) (M^{-1})^{a_1}_{\phantom{a_1}b_1} (M^{-1})^{a_2}_{\phantom{a_2}b_2} \beta^{b_1b_2}.$$ More generally, let $$\textrm{adj}_m(M)^{a_1 \cdots a_m}_{\phantom{a_1 \cdots a_m}{b_1 \cdots b_m}} = \frac{1}{(n-m)!\times m!}\epsilon^{a_1 \cdots a_m i_{m+1}\cdots i_n}\epsilon_{b_1 \cdots b_m j_{m+1}\cdots j_n}M^{j_{m+1}}_{\phantom{j_{m+1}}{i_{m+1}}}\cdots M^{j_n}_{\phantom{j_n}i_n}.\qquad(1)$$ One might perhaps call $\textrm{adj}_m(M)$ the "order-$m$ adjugate matrix of $M$" (although perhaps another accepted name could be found in the literature). It can be shown that for a degree-$m$ fully skew-symmetric tensor $\gamma$ $$\textrm{adj}_m(M)^{a_1 \cdots a_m}_{\phantom{a_1 \cdots a_m}{b_1 \cdots b_m}} \gamma^{b_{1}\cdots b_m} = (\det M) (M^{-1})^{a_1}_{\phantom{a_1}b_1} \cdots (M^{-1})^{a_m}_{\phantom{a_m}b_m} \gamma^{b_1 \cdots b_m}.$$

Proof

In the proof I will refer to some results that are either shown or left as an exercise in the book Differential Geometry and Lie Groups for Physicists by Fecko. To start, multiply equation (1) by $M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m}$ to get $$M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m} \textrm{adj}_m(M)^{a_1 \cdots a_m}_{\phantom{a_1 \cdots a_m}{b_1 \cdots b_m}} = \frac{1}{(n-m)!\times m!}\epsilon^{a_1 \cdots a_m i_{m+1}\cdots i_n}\epsilon_{b_1 \cdots b_m j_{m+1}\cdots j_n}M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m}M^{j_{m+1}}_{\phantom{j_{m+1}}{i_{m+1}}}\cdots M^{j_n}_{\phantom{j_n}i_n}\qquad(2).$$ Observe that on the right-hand side the $n$ superscripts of the first Levi-Civita symbol match precisely the subscripts of the $n$ matrices $M$. We use that $$\epsilon^{a_1 \cdots a_m i_{m+1}\cdots i_n} M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m}M^{j_{m+1}}_{\phantom{j_{m+1}}{i_{m+1}}}\cdots M^{j_n}_{\phantom{j_n}i_n} = (\det M) \epsilon^{c_1 \cdots c_m j_{m+1} \cdots j_n}$$ (this is essentially contained in exercise $\boxed{5.6.5}$ of the book by Fecko). Therefore, equation (2) is rewritten to $$M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m} \textrm{adj}_m(M)^{a_1 \cdots a_m}_{\phantom{a_1 \cdots a_m}{b_1 \cdots b_m}} = \frac{n!}{(n-m)!\times m!}\delta^{c_1 \cdots c_m j_{m+1}\cdots j_n}_{b_1 \cdots b_m j_{m+1}\cdots j_n}(\det M).\qquad (3)$$ where we utilized the generalized Kronecker symbol $$\delta^{a_1\cdots a_n}_{b_1 \cdots b_n} = \frac{1}{n!} \epsilon^{a_1 \cdots a_n}\epsilon_{b_1 \cdots b_n}.$$ Note that $(n-m)$ indices of the generalized Kronecker symbol are contracted in equation (3), so we use $$\delta^{c_1 \cdots c_m j_{m+1}\cdots j_n}_{b_1 \cdots b_m j_{m+1}\cdots j_n} = \frac{1}{\binom{n}{m}} \delta^{c_1 \cdots c_m}_{b_1 \cdots b_m}$$ (exercise $\boxed{5.6.4}$ in Fecko's book). With this, equation (3) simplifies to $$M^{c_1}_{\phantom{c_1}a_1}\cdots M^{c_m}_{\phantom{c_m}a_m} \textrm{adj}_m(M)^{a_1 \cdots a_m}_{\phantom{a_1 \cdots a_m}{b_1 \cdots b_m}} = \delta^{c_1 \cdots c_m}_{b_1 \cdots b_m}(\det M).$$ From here, multiplication with $\gamma^{b_1 \cdots b_m} (M^{-1})^{d_1}_{\phantom{d_1}c_1}\cdots (M^{-1})^{d_m}_{\phantom{d_m}c_m}$ and some basic manipulation with Kronecker deltas trivially gives $$\textrm{adj}_m(M)^{d_1 \cdots d_m}_{\phantom{d_1 \cdots d_m}{b_1 \cdots b_m}} \gamma^{b_{1}\cdots b_m} = (\det M) (M^{-1})^{d_1}_{\phantom{d_1}c_1} \cdots (M^{-1})^{d_m}_{\phantom{d_m}c_m} \gamma^{c_1 \cdots c_m}$$ which (up to some renamed indices) is the identity we set out to prove.

$\endgroup$

You must log in to answer this question.

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