The Green Tent
Column $j$ of $A^{-1}$ is not an abstract strip of floats. It is the steady temperature of a rod with one watt injected at node $j$ and both ends iced to zero — a tent with straight flanks, one kink, and a peak that already knows the suite's variances and resistances. One watt, two walls, one kink.
One watt, two heat baths
Take the rod on $[0,1]$ and clamp both ends to $0^\circ$ — not gently, but with infinite heat baths: whatever heat arrives at an end is swallowed whole, and the end never warms. "Pinned" in the matrix and "perfect heat sink" in the lab are the same boundary condition. Now inject a steady one watt at position $s$ and wait for equilibrium. The temperature profile that remains is the Green's function: $-u''=\delta_s$, $u(0)=u(1)=0$, $u(x) = G(x,s)$.
On the lattice this is a column read. With $n=31$ interior nodes, $h=\tfrac1{32}$, and $A=\mathrm{tridiag}(-1,2,-1)/h^2$ — the suite's stiffness-and-precision matrix — solving $Au = e_j/h$ lands exactly on $G(x_i,s)$ at every node, with $s=x_j$. So column $j$ of $A^{-1}$ is this temperature profile, up to one factor-of-$h$ of lattice normalization that we will pay honest attention to in §6.
One symmetry comes free and is worth saying in physics rather than algebra. $A^{-1}$ is symmetric, so $G(x,s)=G(s,x)$: a watt at $s$ warms $x$ exactly as much as a watt at $x$ warms $s$. That is reciprocity — the same reciprocity Maxwell proved for loaded frames in 1864 — and on this page it is a $31\times31$ float identity with error $10^{-16}$.
Everything about the tent is bookkeeping for one watt
Here is the whole object at $s=0.3125$, with every feature labeled. Four observations, each one line of physics:
(a) The flanks are straight because nothing happens there. Away from the source, steady state means the heat current is constant along the rod — there is nowhere for it to accumulate. Fourier's law says flux $=-u'$, so constant flux is constant slope: a straight line is what "no source here" looks like at equilibrium. The tent rises as $x(1-s)$ on the left of the source and falls as $s(1-x)$ on the right — compactly, $G(x,s)=\min(x,s)-xs$.
(b) The kink is the source. Crossing $s$, the slope drops from $0.6875$ to $-0.3125$: a jump of exactly $-1$. That is $-u''=\delta_s$ read pointwise — the second derivative is a spike of weight one located precisely at the injection, and its size is the wattage. No source, no kink; bigger source, sharper kink.
(c) The slopes are the flux split. The left flank has slope $1-s=0.6875$, so $0.6875$ watts flow out through the left wall; the right flank has slope $-s$, so $0.3125$ watts exit right. The nearer wall drains more. An electrician sees a current divider: from the source there is resistance $s$ to the left wall and $1-s$ to the right, and the unit current splits inversely — $\tfrac{1-s}{1}$ and $\tfrac{s}{1}$. The two exits sum to the watt, to $2\times10^{-15}$.
(d) The peak is a resistance, a variance, and a Green's function diagonal at once. The tent tops out at $G(s,s)=s(1-s)=0.21484375$ — which is exactly the parallel combination $s\,\|\,(1-s)$ of the two paths to ground: the effective resistance from the source to the walls. It is also, times $h$, the Brownian-bridge variance $\Sigma_{jj}$ of report 09, and the quantity the suite's electrical scripts (mathematica/electrical_networks.wls) compute as Vishnoi's effective resistance. One number, three vocabularies. Over all source positions the peaks trace the dashed parabola $s(1-s)$, maxing at $0.25$ mid-rod.
Move the source
Drag $s$ across the rod. The peak rides the gold parabola; the wall fluxes re-divide as $1-s$ and $s$; the kink stays $-1$ — the watt does not care where it stands.
Where the constant diagonal comes from
A fair objection from anyone who has met graph Laplacians: the path graph's Laplacian has diagonal equal to the degrees — $\{1,2,\dots,2,1\}$, with lonely $1$s at the ends — yet the suite's $A$ has a stubbornly constant diagonal of $2$ (over $h^2$), ends included. Which one is "the" discrete $-\mathrm{d}^2/\mathrm{d}x^2$?
Both. $A$ is the free Laplacian of the $(n{+}2)$-vertex path — all $33$ nodes, walls included — with the two pinned end vertices' rows and columns deleted. On the $8$-path this is a two-line computation: strike rows and columns $1$ and $8$ from $L(P_8)$ and the $6\times6$ block that survives is exactly $\mathrm{tridiag}(-1,2,-1)$, constant diagonal and all.
Watch what the deletion does to the end rows. Interior vertex $2$ had degree $2$ and two $-1$ couplings — one to vertex $3$, one to the pin. The pin's column is gone, but the $2$ on the diagonal remembers both neighbors. So the end row now sums to $1$ instead of $0$: the row "leaks". Electrically a row that does not sum to zero is a node with a wire to ground — current can leave the network there. Those two leaks are precisely what evict the constant vector from the kernel: $L\mathbf{1}=0$ made the free Laplacian singular, but $A\mathbf{1}$ has $1/h^2$ poking out at both ends, and $A$ is invertible. Grounding is what buys you an inverse — and hence a Green's function at all.
Statistically the same deletion is conditioning. For a Gaussian field, conditioning on a subset of variables takes a principal submatrix of the precision matrix — that is the whole rule, and it is why precision (not covariance) is the natural home of Markov structure. Pin the two wall values of the free field and the conditional law of the interior has precision $A$: its covariance $A^{-1}=h\,x_i(1-x_j)$, $i\le j$, is the Brownian bridge of report 09. Clamping is conditioning; the number of pins is the number of constraints:
| pins | matrix (over $h^2$) | inverse | process |
|---|---|---|---|
| 0 | free Laplacian, diag $\{1,2,\dots,2,1\}$ | none — singular; $L^{+}$ yields the pairwise effective resistances $R_{ij}=L^{+}_{ii}+L^{+}_{jj}-2L^{+}_{ij}$ | intrinsic field, defined up to a constant |
| 1 | $\mathrm{tridiag}(-1,2,-1)$ with last diagonal $1$ | $h\,\min(x_i,x_j)$ | Brownian motion — the free end wanders, $\mathrm{Var}=h\,x_i$, reaching $0.0302734375$ at $x_{31}$ |
| 2 | $\mathrm{tridiag}(-1,2,-1)$ | $h\,x_i(1-x_j)$, $i\le j$ | Brownian bridge — pinned both ends, variance $h\,s(1-s)$ |
The one-pin row is checked numerically too: flip a single diagonal entry from $2$ to $1$ — free one end — and the inverse switches from the tent product to $h\min(x_i,x_j)$, the Brownian-motion covariance, to $3\times10^{-16}$. One diagonal entry is the difference between a bridge and a random walk.
The same object under many names
The tent has been discovered, renamed, and re-derived for two centuries. Three threads, deliberately told in the vocabulary of each community:
Physics
Before 1828. Newton's $1/r$, then Laplace and Poisson's potential theory, are Green's functions avant la lettre — the free-space kernel before anyone thought to name the boundary-value version. Fourier's 1822 Théorie analytique de la chaleur solves heat problems by superposing elementary solutions: the method is waiting for its kernel.
1828. George Green — a Nottingham miller's son, almost entirely self-taught — publishes his Essay by private subscription, about fifty subscribers. Its move: solve the grounded-conductor electrostatics problem once, for a unit point charge, and every source follows by superposition. The pinned rod is the same mathematics with heat for charge.
1845–54. William Thomson (Kelvin) finds the forgotten essay, and has it republished in Crelle's Journal across 1850–54. The kernel finally gets its audience — and Green's name.
Engineering
The tent is an influence line. Pull a taut string (unit tension, pinned ends) down with a unit point load at $s$: the deflection at $x$ is exactly $x(1-s)$ / $s(1-x)$ — the Green tent, drawn in string. Structural engineers call the load-position-to-response map the influence line.
1864–72. Maxwell's reciprocal theorem for frames — load here, read there, swap, same number — is $G(x,s)=G(s,x)$; Betti generalizes it in 1872. Müller-Breslau (in the 1880s) turns influence lines into the designer's everyday instrument.
20th century. Signals people meet the same kernel as the impulse response of an LTI system: hit it with $\delta$, record $g$, and every input is answered by convolution $u=g*f$. Superposition against a kernel, verbatim.
Quantum
1920s–40s. Quantum mechanics needs the amplitude to go from here to there: the propagator, a Green's function of the Schrödinger (then Dirac) operator.
Late 1940s. Feynman and Schwinger rebuild electrodynamics around it. Every internal line of a Feynman diagram is a Green's function; perturbation theory is superposition of kernels, industrialized. The miller's trick — solve for the point source, sum — becomes the computational engine of QFT.
Superposition is the whole algorithm
Why care so much about one column? Because every right-hand side is a sum of point sources, so every solution is a sum of scaled tents: $u=\sum_j f_j\,G(\cdot,s_j)$ — the discrete shadow of $u(x)=\int_0^1 G(x,y)f(y)\,\mathrm{d}y$. Worked honestly: put a $+2$-watt heater at $s_1=7/32$ and a $-1.5$-watt cooler (a heat pump run backwards — its tent pitched upside-down) at $s_2=25/32$:
And the walls do their own accounting, for free. Each tent ships $f_j(1-s_j)$ watts left and $f_j s_j$ right, so the totals are linear reads of the source list: out the left wall, $2\cdot0.78125-1.5\cdot0.21875 = 1.5625-0.328125 = \mathbf{1.234375}$ W; out the right wall, $0.4375-1.171875=\mathbf{-0.734375}$ W. The negative sign is real physics: the cooler drinks harder than the heater ships rightward, so the right-hand bath pushes $0.734375$ W into the rod. Net: $1.234375-0.734375=0.5$ W leaving — exactly the $2-1.5$ W injected. Steady state balances its books.
Where the suite has already met this tent
As a covariance — report 09: $\Sigma=A^{-1}= h\,x_i(1-x_j)$ for $i\le j$ is the Brownian-bridge covariance, and kriging the interior from boundary data is regression against tent columns. The peak row of this page is its diagonal.
As a response — report 10: the fluctuation–dissipation identity says the deterministic answer to a unit kick, $A^{-1}e_j$, equals the covariance column of the spontaneous thermal jitter. The tent is what you see whether you push the rod or merely watch it tremble.
As a resistance — the electrical-networks scripts (mathematica/electrical_networks.wls, electrical_networks_grid3.wls, after Vishnoi's Lx = b): the peak $s(1-s)$ is effective resistance to ground, the flux split is the current divider, and the grounded rows of §4 are literal wires.
As exact fractions — report 15 runs the $n=5$ chain in rational arithmetic, where the tents' entries are the fractions $\tfrac1{24}$, $\tfrac1{54}$, $\tfrac1{216}$ of the bridge explainer — this page's sibling, which tells the probability half of the story.
As an influence line — the stiffness matrix of §4 is a taut string in matrix clothing, and each column of its inverse is the string's deflection under a unit load: Müller-Breslau for linear algebraists. Superposition of tents is not a metaphor for solving $Au=b$; it is solving $Au=b$.